Module 12: Exponential and Logarithmic Equations and Models

Exponential growth and decay, learning outcomes.

  • Graph exponential growth and decay functions.
  • Solve problems involving radioactive decay, carbon dating, and half life.

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

[latex]y={A}_{0}{e}^{kt}[/latex]

where [latex]{A}_{0}[/latex] is equal to the value at time zero, e  is Euler’s constant, and k  is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time , the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[/latex] where [latex]{A}_{0}[/latex] is the starting value, and e  is Euler’s constant. Now k  is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life , or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the x -axis, they are really a tiny distance above the x -axis.

Graph of y=2e^(3x) with the labeled points (-1/3, 2/e), (0, 2), and (1/3, 2e) and with the asymptote at y=0.

A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[/latex].

Graph of y=3e^(-2x) with the labeled points (-1/2, 3e), (0, 3), and (1/2, 3/e) and with the asymptote at y=0.

A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[/latex].

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri , measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\times {10}^{13}[/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[/latex].

A General Note: Characteristics of the Exponential Function [latex]y=A_{0}e^{kt}[/latex]

An exponential function of the form [latex]y={A}_{0}{e}^{kt}[/latex] has the following characteristics:

  • one-to-one function
  • horizontal asymptote: y  = 0
  • domain: [latex]\left(-\infty , \infty \right)[/latex]
  • range: [latex]\left(0,\infty \right)[/latex]
  • x intercept: none
  • y-intercept: [latex]\left(0,{A}_{0}\right)[/latex]
  • increasing if k  > 0
  • decreasing if k  < 0

An exponential function models exponential growth when k > 0 and exponential decay when k < 0.

Example: Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[/latex] we use the fact that [latex]{A}_{0}[/latex] is the amount at time zero, so [latex]{A}_{0}=10[/latex]. To find k , use the fact that after one hour [latex]\left(t=1\right)[/latex] the population doubles from 10 to 20. The formula is derived as follows:

[latex]\begin{array}{l}\text{ }20=10{e}^{k\cdot 1}\hfill & \hfill \\ \text{ }2={e}^{k}\hfill & \text{Divide both sides by 10}\hfill \\ \mathrm{ln}2=k\hfill & \text{Take the natural logarithm of both sides}\hfill \end{array}[/latex]

so [latex]k=\mathrm{ln}\left(2\right)[/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\left(\mathrm{ln}2\right)t}=10{\left({e}^{\mathrm{ln}2}\right)}^{t}=10\cdot {2}^{t}[/latex]. The graph is shown below.

A graph starting at ten on the y-axis and rising rapidly to the right.

The graph of [latex]y=10{e}^{\left(\mathrm{ln}2\right)t}[/latex] .

Analysis of the Solution

The population of bacteria after ten hours is 10,240. We could describe this amount as being of the order of magnitude [latex]{10}^{4}[/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[/latex], so we could say that the population has increased by three orders of magnitude in ten hours.

Calculating Doubling Time

For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time .

Given the basic exponential growth equation [latex]A={A}_{0}{e}^{kt}[/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[/latex].

The formula is derived as follows:

[latex]\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\hfill & \hfill \\ 2={e}^{kt}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}2=kt\hfill & \text{Take the natural logarithm of both sides}.\hfill \\ t=\frac{\mathrm{ln}2}{k}\hfill & \text{Divide by the coefficient of }t.\hfill \end{array}[/latex]

Thus the doubling time is

[latex]t=\frac{\mathrm{ln}2}{k}[/latex]

Example: Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

[latex]\begin{array}{l}t=\frac{\mathrm{ln}2}{k}\hfill & \text{The doubling time formula}.\hfill \\ 2=\frac{\mathrm{ln}2}{k}\hfill & \text{Use a doubling time of two years}.\hfill \\ k=\frac{\mathrm{ln}2}{2}\hfill & \text{Multiply by }k\text{ and divide by 2}.\hfill \\ A={A}_{0}{e}^{\frac{\mathrm{ln}2}{2}t}\hfill & \text{Substitute }k\text{ into the continuous growth formula}.\hfill \end{array}[/latex]

The function is [latex]A={A}_{0}{e}^{\frac{\mathrm{ln}2}{2}t}[/latex].

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

[latex]f\left(t\right)={A}_{0}{e}^{\frac{\mathrm{ln}2}{3}t}[/latex]

We now turn to exponential decay . One of the common terms associated with exponential decay, as stated above, is half-life , the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

[latex]\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[/latex]

We find that the half-life depends only on the constant k  and not on the starting quantity [latex]{A}_{0}[/latex].

The formula is derived as follows

[latex]\begin{array}{l}\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\hfill & \hfill \\ \frac{1}{2}={e}^{kt}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(\frac{1}{2}\right)=kt\hfill & \text{Take the natural log of both sides}.\hfill \\ -\mathrm{ln}\left(2\right)=kt\hfill & \text{Apply properties of logarithms}.\hfill \\ -\frac{\mathrm{ln}\left(2\right)}{k}=t\hfill & \text{Divide by }k.\hfill \end{array}[/latex]

Since t , the time, is positive, k  must, as expected, be negative. This gives us the half-life formula

[latex]t=-\frac{\mathrm{ln}\left(2\right)}{k}[/latex]

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . The table below lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

[latex]\begin{array}{l}A\left(t\right)={A}_{0}{e}^{\frac{\mathrm{ln}\left(0.5\right)}{T}t}\hfill \\ A\left(t\right)={A}_{0}{e}^{\mathrm{ln}\left(0.5\right)\frac{t}{T}}\hfill \\ A\left(t\right)={A}_{0}{\left({e}^{\mathrm{ln}\left(0.5\right)}\right)}^{\frac{t}{T}}\hfill \\ A\left(t\right)={A}_{0}{\left(\frac{1}{2}\right)}^{\frac{t}{T}}\hfill \end{array}[/latex]

  • [latex]{A}_{0}[/latex] is the amount initially present
  • [latex]T[/latex] is the half-life of the substance
  • [latex]t[/latex] is the time period over which the substance is studied
  • [latex]A[/latex], or [latex]A(t)[/latex], is the amount of the substance present after time [latex]t[/latex]

How To: Given the half-life, find the decay rate

  • Write [latex]A={A}_{o}{e}^{kt}[/latex].
  • Replace A  by [latex]\frac{1}{2}{A}_{0}[/latex] and replace t  by the given half-life.
  • Solve to find k . Express k  as an exact value (do not round).

Note: It is also possible to find the decay rate using [latex]k=-\frac{\mathrm{ln}\left(2\right)}{t}[/latex].

Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for 10% of a 1000-gram sample of uranium-235 to decay?

[latex]\begin{array}{l}\text{ }y=\text{1000}e\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t\hfill & \hfill \\ \text{ }900=1000{e}^{\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t}\hfill & \text{After 10% decays, 900 grams are left}.\hfill \\ \text{ }0.9={e}^{\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t}\hfill & \text{Divide by 1000}.\hfill \\ \mathrm{ln}\left(0.9\right)=\mathrm{ln}\left({e}^{\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t}\right)\hfill & \text{Take ln of both sides}.\hfill \\ \mathrm{ln}\left(0.9\right)=\frac{\mathrm{ln}\left(0.5\right)}{\text{703,800,000}}t\hfill & \text{ln}\left({e}^{M}\right)=M\hfill \\ \text{}\text{}t=\text{703,800,000}\times \frac{\mathrm{ln}\left(0.9\right)}{\mathrm{ln}\left(0.5\right)}\text{years}\hfill & \text{Solve for }t.\hfill \\ \text{}\text{}t\approx \text{106,979,777 years}\hfill & \hfill \end{array}[/latex]

Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

[latex]t=703,800,000\times \frac{\mathrm{ln}\left(0.8\right)}{\mathrm{ln}\left(0.5\right)}\text{ years }\approx \text{ }226,572,993\text{ years}[/latex].

Example: Finding the Function that Describes Radioactive Decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t .

This formula is derived as follows.

[latex]\begin{array}{l}\text{}A={A}_{0}{e}^{kt}\hfill & \text{The continuous growth formula}.\hfill \\ 0.5{A}_{0}={A}_{0}{e}^{k\cdot 5730}\hfill & \text{Substitute the half-life for }t\text{ and }0.5{A}_{0}\text{ for }f\left(t\right).\hfill \\ \text{}0.5={e}^{5730k}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(0.5\right)=5730k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{}k=\frac{\mathrm{ln}\left(0.5\right)}{5730}\hfill & \text{Divide by the coefficient of }k.\hfill \\ \text{}A={A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}\hfill & \text{Substitute for }r\text{ in the continuous growth formula}.\hfill \end{array}[/latex]

The function that describes this continuous decay is [latex]f\left(t\right)={A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}[/latex]. We observe that the coefficient of t , [latex]\frac{\mathrm{ln}\left(0.5\right)}{5730}\approx -1.2097[/latex] is negative, as expected in the case of exponential decay.

The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time measured in years.

[latex]f\left(t\right)={A}_{0}{e}^{-0.0000000087t}[/latex]

Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating  which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12 which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t  years is

[latex]A\approx {A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}[/latex]

  • [latex]A[/latex] is the amount of carbon-14 remaining
  • [latex]{A}_{0}[/latex] is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

[latex]\begin{array}{l}\text{ }A={A}_{0}{e}^{kt}\hfill & \text{The continuous growth formula}.\hfill \\ \text{ }0.5{A}_{0}={A}_{0}{e}^{k\cdot 5730}\hfill & \text{Substitute the half-life for }t\text{ and }0.5{A}_{0}\text{ for }f\left(t\right).\hfill \\ \text{ }0.5={e}^{5730k}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(0.5\right)=5730k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{ }k=\frac{\mathrm{ln}\left(0.5\right)}{5730}\hfill & \text{Divide both sides by the coefficient of }k.\hfill \\ \text{ }A={A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}\hfill & \text{Substitute for }r\text{ in the continuous growth formula}.\hfill \end{array}[/latex]

To find the age of an object we solve this equation for t :

[latex]t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}[/latex]

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r  be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated determined by a method called liquid scintillation. From the equation [latex]A\approx {A}_{0}{e}^{-0.000121t}[/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\frac{A}{{A}_{0}}\approx {e}^{-0.000121t}[/latex]. We solve this equation for t , to get

[latex]t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}[/latex]

How To: Given the percentage of carbon-14 in an object, determine its age

  • Express the given percentage of carbon-14 as an equivalent decimal r .
  • Substitute for r  in the equation [latex]t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}[/latex] and solve for the age, t .

Example: Finding the Age of a Bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

We substitute 20% = 0.20 for r  in the equation and solve for t :

[latex]\begin{array}{l}t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}\hfill & \text{Use the general form of the equation}.\hfill \\ =\frac{\mathrm{ln}\left(0.20\right)}{-0.000121}\hfill & \text{Substitute for }r.\hfill \\ \approx 13301\hfill & \text{Round to the nearest year}.\hfill \end{array}[/latex]

The bone fragment is about 13,301 years old.

The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\text{13,301 years}\pm \text{1% or 13,301 years}\pm \text{133 years}[/latex].

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

less than 230 years; 229.3157 to be exact

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  • 6.7 Exponential and Logarithmic Models
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.3 Radicals and Rational Exponents
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Model exponential growth and decay.
  • Use Newton’s Law of Cooling.
  • Use logistic-growth models.
  • Choose an appropriate model for data.
  • Express an exponential model in base e e .

We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling.

Modeling Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

where A 0 A 0 is equal to the value at time zero, e e is Euler’s constant, and k k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time , the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form y = A 0 e k t y = A 0 e k t where A 0 A 0 is the starting value, and e e is Euler’s constant. Now k k is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life , or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in Figure 2 and Figure 3 . It is important to remember that, although parts of each of the two graphs seem to lie on the x -axis, they are really a tiny distance above the x -axis.

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri , measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 × 10 13 . 4.01134972 × 10 13 . So, we could describe this number as having order of magnitude 10 13 . 10 13 .

Characteristics of the Exponential Function, y = A 0 e k t y = A 0 e k t

An exponential function with the form y = A 0 e k t y = A 0 e k t has the following characteristics:

  • one-to-one function
  • horizontal asymptote: y = 0 y = 0
  • domain: ( – ∞ ,   ∞ ) ( – ∞ ,   ∞ )
  • range: ( 0 , ∞ ) ( 0 , ∞ )
  • x intercept: none
  • y-intercept: ( 0 , A 0 ) ( 0 , A 0 )
  • increasing if k > 0 k > 0 (see Figure 4 )
  • decreasing if k < 0 k < 0 (see Figure 4 )

Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find A 0 A 0 we use the fact that A 0 A 0 is the amount at time zero, so A 0 = 10. A 0 = 10. To find k , k , use the fact that after one hour ( t = 1 ) ( t = 1 ) the population doubles from 10 10 to 20. 20. The formula is derived as follows

so k = ln ( 2 ) . k = ln ( 2 ) . Thus the equation we want to graph is y = 10 e ( ln 2 ) t = 10 ( e ln 2 ) t = 10 · 2 t . y = 10 e ( ln 2 ) t = 10 ( e ln 2 ) t = 10 · 2 t . The graph is shown in Figure 5 .

The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude 10 4 . 10 4 . The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude 10 7 , 10 7 , so we could say that the population has increased by three orders of magnitude in ten hours.

We now turn to exponential decay . One of the common terms associated with exponential decay, as stated above, is half-life , the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

We find that the half-life depends only on the constant k k and not on the starting quantity A 0 . A 0 .

The formula is derived as follows

Since t , t , the time, is positive, k k must, as expected, be negative. This gives us the half-life formula

Given the half-life, find the decay rate.

  • Write A = A o e k t . A = A o e k t .
  • Replace A A by 1 2 A 0 1 2 A 0 and replace t t by the given half-life.
  • Solve to find k . k . Express k k as an exact value (do not round).

Note: It is also possible to find the decay rate using k = − ln ( 2 ) t . k = − ln ( 2 ) t .

Finding the Function that Describes Radioactive Decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t . t .

This formula is derived as follows.

The function that describes this continuous decay is f ( t ) = A 0 e ( ln ( 0.5 ) 5730 ) t . f ( t ) = A 0 e ( ln ( 0.5 ) 5730 ) t . We observe that the coefficient of t , t , ln ( 0.5 ) 5730 ≈ − 1.2097 × 10 −4 ln ( 0.5 ) 5730 ≈ − 1.2097 × 10 −4 is negative, as expected in the case of exponential decay.

The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time, measured in years.

Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating , which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t t years is

  • A A is the amount of carbon-14 remaining
  • A 0 A 0 is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

To find the age of an object, we solve this equation for t : t :

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r r be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation A ≈ A 0 e − 0.000121 t A ≈ A 0 e − 0.000121 t we know the ratio of the percentage of carbon-14 in the object we are dating to the initial amount of carbon-14 in the object when it was formed is r = A A 0 ≈ e − 0.000121 t . r = A A 0 ≈ e − 0.000121 t . We solve this equation for t , t , to get

Given the percentage of carbon-14 in an object, determine its age.

  • Express the given percentage of carbon-14 as an equivalent decimal, k . k .
  • Substitute for k in the equation t = ln ( r ) − 0.000121 t = ln ( r ) − 0.000121 and solve for the age, t . t .

Finding the Age of a Bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

We substitute 20 % = 0.20 20 % = 0.20 for r r in the equation and solve for t : t :

The bone fragment is about 13,301 years old.

The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as 13,301 years ± 1% or 13,301 years ± 133 years . 13,301 years ± 1% or 13,301 years ± 133 years .

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

Calculating Doubling Time

For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time .

Given the basic exponential growth equation A = A 0 e k t , A = A 0 e k t , doubling time can be found by solving for when the original quantity has doubled, that is, by solving 2 A 0 = A 0 e k t . 2 A 0 = A 0 e k t .

The formula is derived as follows:

Thus the doubling time is

Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

The function is A 0 e ln 2 2 t . A 0 e ln 2 2 t .

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

Using Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling , the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature

  • Newton’s Law of Cooling

The temperature of an object, T , T , in surrounding air with temperature T s T s will behave according to the formula

  • t t is time
  • A A is the difference between the initial temperature of the object and the surroundings
  • k k is a constant, the continuous rate of cooling of the object

Given a set of conditions, apply Newton’s Law of Cooling.

  • Set T s T s equal to the y -coordinate of the horizontal asymptote (usually the ambient temperature).
  • Substitute the given values into the continuous growth formula T ( t ) = A e k t + T s T ( t ) = A e k t + T s to find the parameters A A and k . k .
  • Substitute in the desired time to find the temperature or the desired temperature to find the time.

A cheesecake is taken out of the oven with an ideal internal temperature of 165°F, 165°F, and is placed into a 35°F 35°F refrigerator. After 10 minutes, the cheesecake has cooled to 150°F . 150°F . If we must wait until the cheesecake has cooled to 70°F 70°F before we eat it, how long will we have to wait?

Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation

We know the initial temperature was 165, so T ( 0 ) = 1 6 5 . T ( 0 ) = 1 6 5 .

We were given another data point, T ( 1 0 ) = 1 5 0 , T ( 1 0 ) = 1 5 0 , which we can use to solve for k . k .

This gives us the equation for the cooling of the cheesecake: T ( t ) = 1 3 0 e – 0 . 0 1 2 3 t + 3 5 . T ( t ) = 1 3 0 e – 0 . 0 1 2 3 t + 3 5 .

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70°F . 70°F .

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?

Using Logistic Growth Models

Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity . For constants a, b, a, b, and c, c, the logistic growth of a population over time t t is represented by the model

The graph in Figure 6 shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.

Logistic Growth

The logistic growth model is

  • c 1 + a c 1 + a is the initial value
  • c c is the carrying capacity , or limiting value
  • b b is a constant determined by the rate of growth.

Using the Logistic-Growth Model

An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.

For example, at time t = 0 t = 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is b = 0.6030. b = 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

We substitute the given data into the logistic growth model

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is c = 1000. c = 1000. To find a , a , we use the formula that the number of cases at time t = 0 t = 0 is c 1 + a = 1 , c 1 + a = 1 , from which it follows that a = 999. a = 999. This model predicts that, after ten days, the number of people who have had the flu is f ( t ) = 1000 1 + 999 e − 0.6030 x ≈ 293.8. f ( t ) = 1000 1 + 999 e − 0.6030 x ≈ 293.8. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, c = 1000. c = 1000.

Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.

The graph in Figure 7 gives a good picture of how this model fits the data.

Using the model in Example 6 , estimate the number of cases of flu on day 15.

Choosing an Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed in Table 1 ? Find the model, and use a graph to check your choice.

First, plot the data on a graph as in Figure 8 . For the purpose of graphing, round the data to two decimal places.

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try y = a ln ( b x ) . y = a ln ( b x ) . Plugging in the first point, ( 1,0 ) , ( 1,0 ) , gives 0 = a ln b . 0 = a ln b . We reject the case that a = 0 a = 0 (if it were, all outputs would be 0), so we know ln ( b ) = 0. ln ( b ) = 0. Thus b = 1 b = 1 and y = a ln ( x ) . y = a ln ( x ) . Next we can use the point ( 9,4 .394 ) ( 9,4 .394 ) to solve for a : a :

Because a = 4.394 ln ( 9 ) ≈ 2 , a = 4.394 ln ( 9 ) ≈ 2 , an appropriate model for the data is y = 2 ln ( x ) . y = 2 ln ( x ) .

To check the accuracy of the model, we graph the function together with the given points as in Figure 9 .

We can conclude that the model is a good fit to the data.

Compare Figure 9 to the graph of y = ln ( x 2 ) y = ln ( x 2 ) shown in Figure 10 .

The graphs appear to be identical when x > 0. x > 0. A quick check confirms this conclusion: y = ln ( x 2 ) = 2 ln ( x ) y = ln ( x 2 ) = 2 ln ( x ) for x > 0. x > 0.

However, if x < 0 , x < 0 , the graph of y = ln ( x 2 ) y = ln ( x 2 ) includes a “extra” branch, as shown in Figure 11 . This occurs because, while y = 2 ln ( x ) y = 2 ln ( x ) cannot have negative values in the domain (as such values would force the argument to be negative), the function y = ln ( x 2 ) y = ln ( x 2 ) can have negative domain values.

Does a linear, exponential, or logarithmic model best fit the data in Table 2 ? Find the model.

Expressing an Exponential Model in Base e e

While powers and logarithms of any base can be used in modeling, the two most common bases are 10 10 and e . e . In science and mathematics, the base e e is often preferred. We can use laws of exponents and laws of logarithms to change any base to base e . e .

Given a model with the form y = a b x , y = a b x , change it to the form y = A 0 e k x . y = A 0 e k x .

  • Rewrite y = a b x y = a b x as y = a e ln ( b x ) . y = a e ln ( b x ) .
  • Use the power rule of logarithms to rewrite y as y = a e x ln ( b ) = a e ln ( b ) x . y = a e x ln ( b ) = a e ln ( b ) x .
  • Note that a = A 0 a = A 0 and k = ln ( b ) k = ln ( b ) in the equation y = A 0 e k x . y = A 0 e k x .

Changing to base e

Change the function y = 2.5 ( 3.1 ) x y = 2.5 ( 3.1 ) x so that this same function is written in the form y = A 0 e k x . y = A 0 e k x .

Change the function y = 3 ( 0.5 ) x y = 3 ( 0.5 ) x to one having e e as the base.

Access these online resources for additional instruction and practice with exponential and logarithmic models.

  • Logarithm Application – pH
  • Exponential Model – Age Using Half-Life
  • Exponential Growth Given Doubling Time
  • Exponential Growth – Find Initial Amount Given Doubling Time

6.7 Section Exercises

With what kind of exponential model would half-life be associated? What role does half-life play in these models?

What is carbon dating? Why does it work? Give an example in which carbon dating would be useful.

With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?

Define Newton’s Law of Cooling. Then name at least three real-world situations where Newton’s Law of Cooling would be applied.

What is an order of magnitude? Why are orders of magnitude useful? Give an example to explain.

The temperature of an object in degrees Fahrenheit after t minutes is represented by the equation T ( t ) = 68 e − 0.0174 t + 72. T ( t ) = 68 e − 0.0174 t + 72. To the nearest degree, what is the temperature of the object after one and a half hours?

For the following exercises, use the logistic growth model f ( x ) = 150 1 + 8 e − 2 x . f ( x ) = 150 1 + 8 e − 2 x .

Find and interpret f ( 0 ) . f ( 0 ) . Round to the nearest tenth.

Find and interpret f ( 4 ) . f ( 4 ) . Round to the nearest tenth.

Find the carrying capacity.

Graph the model.

Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.

Rewrite f ( x ) = 1.68 ( 0.65 ) x f ( x ) = 1.68 ( 0.65 ) x as an exponential equation with base e e to five decimal places.

For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic.

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in t t years is modeled by the equation P ( t ) = 1000 1 + 9 e − 0.6 t . P ( t ) = 1000 1 + 9 e − 0.6 t .

Graph the function.

What is the initial population of fish?

To the nearest tenth, what is the doubling time for the fish population?

To the nearest whole number, what will the fish population be after 2 2 years?

To the nearest tenth, how long will it take for the population to reach 900 ? 900 ?

What is the carrying capacity for the fish population? Justify your answer using the graph of P . P .

A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?

The formula for an increasing population is given by P ( t ) = P 0 e r t P ( t ) = P 0 e r t where P 0 P 0 is the initial population and r > 0. r > 0. Derive a general formula for the time t it takes for the population to increase by a factor of M .

Recall the formula for calculating the magnitude of an earthquake, M = 2 3 log ( S S 0 ) . M = 2 3 log ( S S 0 ) . Show each step for solving this equation algebraically for the seismic moment S . S .

What is the y -intercept of the logistic growth model y = c 1 + a e − r x ? y = c 1 + a e − r x ? Show the steps for calculation. What does this point tell us about the population?

Prove that b x = e x ln ( b ) b x = e x ln ( b ) for positive b ≠ 1. b ≠ 1.

Real-World Applications

For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.

To the nearest hour, what is the half-life of the drug?

Write an exponential model representing the amount of the drug remaining in the patient’s system after t t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram.

Using the model found in the previous exercise, find f ( 10 ) f ( 10 ) and interpret the result. Round to the nearest hundredth.

For the following exercises, use this scenario: A tumor is injected with 0.5 0.5 grams of Iodine-125, which has a decay rate of 1.15% 1.15% per day.

To the nearest day, how long will it take for half of the Iodine-125 to decay?

Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram.

A scientist begins with 250 250 grams of a radioactive substance. After 250 250 minutes, the sample has decayed to 32 32 grams. Rounding to five decimal places, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?

The half-life of Radium-226 is 1590 1590 years. What is the annual decay rate? Express the decimal result to four decimal places and the percentage to two decimal places.

The half-life of Erbium-165 is 10 .4 10 .4 hours. What is the hourly decay rate? Express the decimal result to four decimal places and the percentage to two decimal places.

A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 573 0 573 0 years.)

A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 1350 bacteria. Rounding to five decimal places, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 3 hours?

For the following exercises, use this scenario: A biologist recorded a count of 360 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes.

To the nearest whole number, what was the initial population in the culture?

Rounding to six decimal places, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?

For the following exercises, use this scenario: A pot of warm soup with an internal temperature of 100° 100° Fahrenheit was taken off the stove to cool in a 69° F 69° F room. After fifteen minutes, the internal temperature of the soup was 95° F . 95° F .

Use Newton’s Law of Cooling to write a formula that models this situation.

To the nearest minute, how long will it take the soup to cool to 80° F? 80° F?

To the nearest degree, what will the temperature be after 2 2 and a half hours?

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of 165°F 165°F and is allowed to cool in a 75°F 75°F room. After half an hour, the internal temperature of the turkey is 145°F . 145°F .

Write a formula that models this situation.

To the nearest degree, what will the temperature be after 50 minutes?

To the nearest minute, how long will it take the turkey to cool to 110° F? 110° F?

For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.

Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper: 10 − 10   W m 2 , 10 − 10   W m 2 , Vacuum: 10 − 4 W m 2 , 10 − 4 W m 2 , Jet: 10 2   W m 2 10 2   W m 2

Recall the formula for calculating the magnitude of an earthquake, M = 2 3 log ( S S 0 ) . M = 2 3 log ( S S 0 ) . One earthquake has magnitude 3 . 9 3 . 9 on the MMS scale. If a second earthquake has 75 0 75 0 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.

For the following exercises, use this scenario: The equation N ( t ) = 500 1 + 49 e − 0.7 t N ( t ) = 500 1 + 49 e − 0.7 t models the number of people in a town who have heard a rumor after t days.

How many people started the rumor?

To the nearest whole number, how many people will have heard the rumor after 3 days?

As t t increases without bound, what value does N ( t ) N ( t ) approach? Interpret your answer.

For the following exercise, choose the correct answer choice.

A doctor injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation?

  • ⓐ f ( t ) = 13 ( 0.0805 ) t f ( t ) = 13 ( 0.0805 ) t
  • ⓑ f ( t ) = 13 e 0.9195 t f ( t ) = 13 e 0.9195 t
  • ⓒ f ( t ) = 13 e ( − 0.0839 t ) f ( t ) = 13 e ( − 0.0839 t )
  • ⓓ f ( t ) = 4.75 1 + 13 e − 0.83925 t f ( t ) = 4.75 1 + 13 e − 0.83925 t

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Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Authors: Jay Abramson
  • Publisher/website: OpenStax
  • Book title: College Algebra 2e
  • Publication date: Dec 21, 2021
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
  • Section URL: https://openstax.org/books/college-algebra-2e/pages/6-7-exponential-and-logarithmic-models

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Exponential-Decay Word Problems

Log Probs Expo Growth Expo Decay

What is the formula for exponential decay?

Exponential growth word problems work off the exponential-decay formula, A  =  Pe kt , where A is the ending amount of whatever you're dealing with (for example, carbon-14 in a biological sample), P is the beginning amount of that same whatever, k is the decay constant, and t is time.

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What is the difference between the exponential-growth formula and the exponential-decay formula?

The only difference between exponential-growth equations and exponential-decay equations is that the constant for decay situations is negative. The equation itself is just the same as for exponential growth, but you should expect a negative value for the constant. If you get a positive value, you should probably go back and check your work.

Note that the particular variables used in the equation may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:

A = P e r t A = P e k t Q = N e k t Q = Q 0 e k t

No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the decay constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

To solve exponential decay word problems, you may be plugging one value into the exponential-decay equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the half-life information to find the decay constant (probably by solving the exponential equation by using logarithms), and then using that value to find whatever the exercise requested.

What is an example of a radioactive-decay exercise?

  • Radio-isotopes of different elements have different half-lives. Magnesium- 27 has a half-life of 9.45 minutes. What is the decay constant for Magnesium- 27 ? Round to five decimal places.

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Since the decay rate is given in terms of minutes, then time t will be in minutes. However, I note that there is no beginning or ending amount given. How am I supposed to figure out what the decay constant is?

I can do this by working from the definition of "half-life": in the given amount of time (in this case, 9.45 minutes), half of the initial amount will be gone. That is, from t  = 0 to t  = 9.45 , I will have gone from 100% of however much I'd started with (where " 100% " is written as 1 , being its decimal equivalent) to 50% of that amount (converted to 0.5 for use inside the formula).

Since the half-life does not depend on how much I started with, I can either pick an arbitrary beginning amount (such as 100 grams) and then calculate the decay constant after 9.45 minutes, at which point only 50 grams will remain (the other 50 grams will have mutated into some other isotope or element). Or else I can just deal with the 50% that is left. Either way, I will end up with this equation:

0.5 = e 9.45 k

Solving for the decay constant k , I get:

ln(0.5) = 9.45 k

ln(0.5) / 9.45 = k = −0.073348907996...

They want me to round to five decimal places, so my answer is:

k = −0.07335

The constant was negative, as expected, because this was a decay problem. If I'd ended up with a positive value, this would have signalled to me that I'd made a mistake somewhere.

(Note: Technically, there should be units on the constant, so that the units on the time variable t will cancel off. In other words, technically, the answer should be " k  = −0.07335 /minute. But this probably won't matter in your math class.)

  • Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. A hepatobiliary scan of my gallbladder involved an injection of 0.5 cc's (or about one-tenth of a teaspoon) of Technetium-99m, which has a half-life of almost exactly 6 hours. While undergoing the test, I heard the technician telling somebody on the phone that "in twenty-four hours, you'll be down to background radiation levels." Figure out just how much radioactive material remained from my gallbladder scan after twenty-four hours. Show your work below.

For this exercise, I need to find the ending amount A of Technetium-99m. Recalling that 1 cc (cubic centimeter) equals 1 mL (milliliter), I know that the beginning amount is P = 0.5 mL. The ending time is 24 hours. I do not have the decay constant but, by using the half-life information, I can find it. (Since this is a decay problem, I expect the constant to be negative. If I end up with a positive value, I'll know that I should go back and check my work.)

In 6 hours, there will be 50% of the original amount left, so:

0.5 = e 6 k

ln(0.5) = 6 k

ln(0.5)/6 = k

(This evaluates to about −0.1155 , but I'll leave the decay constant in exact form to avoid round-off error.)

Now that I have the decay constant, I can find out how much Technitium- 99 m was left after twenty-four hours:

A = 0.5 e (ln(0.5)/6)(24) = 0.03125

There will be no more than 0.03125 mL (or about 1/160 of a teaspoon) of Technitium- 99 m remaining after twenty-four hours.

It can be very useful, sometimes in unexpected places, to know that 1 cc (that is, one cubic centimeter) is equal to 1 mL (that is, one milliliter). I would recommend that you add this bit of knowledge to your store.

Algebra Tutors

By the way: Technetium-99m is one of the most commonly used radioisotope for these medical purposes. Its radiation is extremely low-energy, so the chance of mutation is very low. (Whatever you're being treated for is the greater danger.) The half-life is just long enough for the doctors to have time to take their pictures. The dose I was given is about as large as these injections typically get. Your body does not easily absorb this chemical, so most of the injection is... "voided", shall we say? into the sewer system. 🚽

What is an example of solving a carbon-dating exercise?

  • Carbon- 14 has a half-life of 5730 years. You are presented with a document which purports to contain the recollections of a Mycenaean soldier during the Trojan War. The city of Troy was finally destroyed in about 1250 BC, or about 3250 years ago. Carbon-dating evaluates the ratio of radioactive carbon- 14 to stable carbon- 12 . Given the amount of carbon- 12 contained a measured sample cut from the document, there would have been about 1.3 × 10 −12 grams of carbon- 14 in the sample when the parchment was new, assuming the proposed age is correct. According to your equipment, there remains 1.0 × 10 −12 grams. Is there a possibility that this is a genuine document? Or is this instead a recent forgery? Justify your conclusions.

First, as usual, I have to find the decay rate. (In real life, you'd look this up on a table, or have it programmed into your equipment, but this is math, not real life.) The half-life is 5730 years, so:

0.5 = e 5730 k

ln(0.5) = 5730 k

ln(0.5) / 5730 = k

I'll leave the decay constant in this "exact" form to avoid round-off error.

I have the beginning (expected) amount of C- 14 and the present (ending) amount; from this information, I can calculate the age of the parchment:

1.0 × 10 −12 = (1.3 × 10 −12 ) e (ln(0.5)/5730) t

1 / 1.3 = e (ln(0.5)/5730) t

ln( 1 / 1.3 ) = ( ln(0.5) / 5730 ) t

5730ln( 1 / 1.3 ) / ln(0.5) = t = 2168.87160124...

Then the parchment is about 2170 years old, much less than the necessary 3250 years ago that the Trojan War took place. But the parchment is indeed old, so this isn't a total fake.

Since the parchment is genuinely old ( 2170 years), but clearly not old enough to be the actual writings of a soldier in the Trojan War ( 3250 years), either this is a much-younger copy of an earlier document (in which case it is odd that there are no references to it in other documents, since only famous works tended to be copied), or, which is more likely, this is a recent forgery written on a not-quite-old-enough ancient parchment. If possible, the ink should be tested, since a recent forgery would use recently-made ink.

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how to do half life problems college algebra

Half Life Formula

The half-life formula is used to find the half-life of a substance that is decaying or reducing in quantity. A substance that is decaying has a different rate of decay for different quantities of the substance. As the quantity of the substance reduces the rate of decay also slows down, and hence it is very difficult to find the life of a decaying substance. Therefore the half-life formula is used to provide the right metrics to define the life of decaying material. In this section, let us learn more about the half life formula and solve a few examples .

What is Half Life Formula?

Half-life refers to the amount of time it takes for half of a particular sample to react i.e it refers to the time that a particular quantity requires to reduce its initial value to half. The half-life formula is commonly used in nuclear physics where it describes the speed at which an atom undergoes radioactive decay. The formula for the half-life is obtained by dividing  0.693 by the constant  λ. Here λ is called the disintegration or decay constant. Hence the formula to calculate the half-life of a substance is:

\(t_{\frac{1}{2}} = \dfrac{0.693}{\lambda} \)

\(t_{\frac{1}{2}}\) = half-life

λ = constant

Let N be the size of the population of the radioactive atoms at a given time t, dN be the amount by which it decreases in time dt. The rate of change is given as dN/d t = -λN, where λ is the decay constant.

On integrating this equation, we get N = \(N_0\)e -λt  , where \(N_0\) = the size of the initial population of radioactive atoms at t = 0. This shows that the population decays exponentially at a rate that depends on λ. The time taken for half of the original population of radioactive atoms to decay is called the half-life. This relationship between half-life, the time period, t 1/2 , and the decay constant λ is given by   \(t_{\frac{1}{2}} = \dfrac{0.693}{\lambda} \).

how to do half life problems college algebra

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Examples Using Half Life Formula

Example 1: The decay constant of a substance is 0.84 s -1 . Find the half life of the substance.

Given decay constant λ = 0.84  The half life formula can be used to find the half life of the substance. \(t_{\frac{1}{2}}\)   = 0.693/ λ

= 0.693/0.84

= 0.825 Therefore, the half life of the substance is 0.8 seconds.

Example 2: Find the value of the decay constant of a radioactive substance having a half-life of 0.04 seconds.

Solution: Given half life of the substance is \(t_{\frac{1}{2}}\)= 0.04 The half life formula can be used to find the half life of the substance. \(t_{\frac{1}{2}}\)   = 0.693/ λ

λ = 0.693/0.04

= 17.325 Therefore, the decay constant of the radioactive substance is 17.325 s -1 .

Example 3: Calculate the half-life of a radioactive substance whose disintegration constant happens to be 0.004 1/years?

The quantities available here are,

λ = 0.004  1/years

Using the half life formula, 

\(t_{\frac{1}{2}}\)   = 0.693/ λ

=0.693/0.004

Hence, the half-life of this particular radioactive substance is 173.25 years.

FAQs on Half Life Formula

What is meant by half life formula .

Half-life refers to the amount of time it takes for half of a particular sample to react i.e it refers to the time that a particular quantity requires to reduce its initial value to half. The half-life formula is commonly used in nuclear physics where it describes the speed at which an atom undergoes radioactive decay. The formula is \(t_{\frac{1}{2}}\)= 0.693/ λ

What is the Formula to Find the Half Life of a Substance?

 The formula for the half-life is obtained by dividing 0.693 by the constant λ . Here λ is called the disintegration or decay constant. Hence the formula to calculate the half-life of a value is:

Are the Number Positive or Negative in the Half Life Formula?

Both the time and  λ are positive numbers, where the time shows the time taken for decaying quantity and the  λ is the decay constant of the decaying quantity. 

Using the Half Life Formula, Find the Half Life of a Substance with Disintegration Constant happens to be 0.008 1/years?

λ = 0.008  1/years

= 0.693/0.008

Hence, the half-life of this particular radioactive substance is 86.625 years.

Quantitative Reasoning

Half-Life & Doubling-Time

Exponential growth in terms of doubling.

It is often helpful to talk about exponential growth in terms of "doubling" since this provides an intuitive sense for how the quantity changes over time. If we talk about the population of a village doubling over a decade, or the value of an investment doubling over the course of a few years, we can readily imagine what is being described. Mathematically, doubling is just a special case of the exponential function introduced earlier, where the growth factor, \(B\) is \(2\), and the independent variable is the number of doublings, \(n\).

\(f\left(n\right)=A\cdot{(2)}^{n}\)

Imagine the population in a small town doubles every twelve years. Observe how the output values in the table below change as the input increases by 1. Each output value is the product of the previous output and the base of 2 (it doubles). In this example, we will let the initial population, \(A\), be \(300\).

Looking at the table, we see that after two doublings (24 years), the population has doubled twice, and our exponential function would have an exponent of 2. After three doublings (36 years), the population has doubled three times, and our exponential function would have an exponent of 3.

A small town has an initial population of 300 people. The population doubles every twelve years.

  • What exponent would be used to calculate the population after 48 years (four doublings)?
  • What exponent would be used to calculate the population after 60 years (five doublings)?
  • What exponent would be used to calculate the population after 54 years (four and a half doublings)?
  • What exponent would be used to calculate the population after 51 years (four and a quarter doublings)?
  • What exponent would be used to calculate the population after 49 years?
  • What exponent would be used to calculate the population after \(t\) years?

Show Answer

  • After four doublings (48 years), the exponent would be 4
  • After five doublings (60 years), the exponent would be 5
  • After four and a half doublings (54 years), the exponent would be 5.5
  • After four and a quarter doublings (51 years), the exponent would be 5.25
  • Twelve goes into 49 four times, with a remainder of one, so 49 years would be four doublings plus one twelfth of a doubling. The exponent would be \(4\frac{1}{12}\) or approximately 4.083
  • To find the exponent after \(t\) years, we need to find how many times 12 goes into \(t\). The exponent would just be \(\frac{t}{12}\)

Based on the results of the last example, we can rewrite our population data table and include a column for the year.

Every twelve years, the population doubles, and the exponent becomes an integer based on \(n=t/12\). To model the population at any arbitrary year, we rewrite the exponential function in terms of the doubling time.

or more generally

\(f(t)=300\cdot 2^{(t/T)}\), where \(T\) is the doubling time.

Exponential Growth

A function that models exponential growth doubles in size after a characteristic time, \(T\), called the doubling time . The exponential growth function can be written in the form

  • \(A\) is the initial or starting value of the function.
  • \(t\) is the time that has passed since the growth began.
  • \(T\) is the doubling time : the amount of time required for the function to double in size.
  • \(t/T\) is the ratio describing the number of doublings that have occurred.

A field has an initial population of 8 bunnies. Bunnies reproduce very quickly, and a reasonable estimate is that the bunny population doubles five times every year. In other words, \(T=0.2\) years.

Assuming there are no predators to reduce the population, how many bunnies will be in the field after four years?

We can use the doubling form of the exponential function where our initial value is 8 and our doubling time is 0.2:

To find the population after four years, we substitute \(4\) for \(t\):

\(f(4)=8\cdot 2^{(4/0.2)} = 8\cdot2^{20}=8,388,608\) bunnies.

After four years, the population will have grown from 8 bunnies to nearly 8.4 million bunnies.

When our starting value, \(A\) is repeatedly multiplied by a factor that is greater than 1, the result grows over time. If we know the growth factor per unit time, \(B\), then we can write

\(f(t)=A\cdot(B)^t\), where \(B\gt 1\).

If we know how long it takes for the value to double, we can describe the exact same growth model as

\(f(t)=A\cdot(2)^{t/T}\), where \(T\) is the time needed to double and \(t/T\) is the number of doublings.

Exponential Decay in terms of Half-Life

Exponential decay is the same as exponential growth except we repeatedly multiply by a factor that is between 0 and 1, so the result shrinks over time. If we know the decay factor per unit time, \(B\), then we can write

\(f(t)=A\cdot(B)^t\), where \(0\lt B\lt 1\).

If we know how long it takes for the value to be cut in half, we can describe the exact same growth model as

\(f(t)=A\cdot\left( \frac{1}{2}\right)^{t/T}\), where \(T\) is the time needed for the output value to be cut in half.

We measured exponential growth using the "doubling-time", and we can measure exponential decay using the "halving-time." Exponential decay occurs in many situations in physics, chemistry, engineering, and finance. In situations involving radioactive decay, the constant \(T\) is known as the half-life because it measures the lifetime of a radioactively unstable atom. In fact, the term half-life is often preferred over the more awkward "halving-time" even in situations that have nothing to do with lifetimes.

We will investigate exponential decay in terms of temperature. " Newton's Law of Cooling " can be summed up by saying that heat flows swiftly when a very hot object is placed in a very cold environment, but heat flows slowly when the object is only slightly warmer than the environment. When an object is at the same temperature as its environment, no heat flows at all. When objects cool down, their temperature decays exponentially.

Leaving food out too long at room temperature can cause harmful bacteria to grow to dangerous levels that can cause illness. Bacteria grow most rapidly in the range of temperatures between 40°F and 140°F. This range of temperatures is often called the "Danger Zone." Restaurants that prepare food in advance must cool the food below the danger zone in less than six hours. Simply placing a container of hot soup into a refrigerator will cool the food too slowly, so restaurants typically use ice-water baths before refrigeration.

A large container of 140°F soup is placed in an ice-water bath with a temperature of 0°F. If the "half-life" of the exponential decay is 3.5 hours, will the soup cool below the Danger Zone in less than six hours?

The cooling period of six hours is almost long enough for two "half-lifes" of 3.5 hours each, which would reduce the temperature from 140°F to 70°F to 35°F. So the soup might be safe. To know for sure, we use the exponential decay formula

\(f(6)=140\cdot (0.5)^{6/3.5}=42.7\)°F.

No, the soup will not cool in less than six hours. The proposed cooling method is not safe because the exponential decay is too slow. One solution would be to split the soup into several smaller containers that would cool more swiftly.

Approximate Half-Life and Doubling-Time

Sometimes we know the percent change, \(r\), in \(f(t)=A\cdot(1+r)^t\), but we're interested in the half-life or doubling time. We know that there is an equivalent growth function \(f(t)=A\cdot(2)^{t/T}\) or \(f(t)=A\cdot(1/2)^{t/T}\), but we don't know how to find \(T\) when given \(r\).

There is a handy approximation that relates the percent change to the characteristic time. First, express your percent change as a percentage without the percent sign. For example, if \(r=3.5\%\), then we would say that \(R=3.5\). Just 3.5, not 0.035. In that case, we find: \(70/R\approx T\) where \(T\) is the doubling time or the half-life, and \(R\) is the percent change without the percent sign. The result is only approximate, but the approximation is more accurate when the rate is low. Even for high growth rates (like \(r\gt5\%\)), the approximation can still provide a good starting point for guess & check refinement.

We have seen equations with this form before (remember margin of error?), so we know that it can also be written in a different form: \(70/T\approx R\)

Also note that the time units must match. If your percent change is 20% per minute, then your doubling time will be measured in minutes. If your doubling time is in days, then your change will be a percent change per day.

Exact Half-Life and Doubling-Time using Logarithms or Guess & Check

Gloved hands holding a dish of highly enrich uranium metal.

In some real-world situations, the unknown is in an exponent. For example, we might need to find out the growth rate when given population values over time, or we might need to find the half-life as a quantity decays. Those values are in the exponent, and we only have two ways to find them:

1) We could guess possible values for the exponent, and then check our calculation against the data. We then adjust our guess to improve our accuracy, and repeat until our value is sufficiently accurate. Many students dismiss "guess & check", but it is a fast and easy method, that is also understandable by others. This makes it an excellent solution strategy, particularly when coupled with the \(70/R\) approximation that provides a first guess that is very close to the exact value.

2) It is also possible to use logarithms to find exponent values. This can sometimes be faster, but it is somewhat difficult to understand, and it's certainly more difficult to communicate to others. Logarithms are not required in this course. The following cartoon introduces the concept of the logarithm.

Every logarithm has a base, which is usually written as a subscript (exceptions would be "LOG" which has a base of 10 and no subscript, and "ln" which has a base of e and no subscript). The logarithm acts on a number, converting it into exponential form using that base. The logarithm then gives you the exponent. For example, \(\log_3(81)\) is read as the "log, base 3, of 81". Since the base of the logarithm is 3, the number 81 is converted into exponential form with a base of 3: \(\log_3(3^4)\). The logarithm then converts that into \(4 \cdot \log_3(3)\). Whenever the bases match, the logarithm has a value of 1, so \(4 \cdot \log_3(3) = 4 \cdot 1\). In short, \(\log_3(81)=4\).

The base of the logarithm is very important. For example, \(\log_9(81) = 2\) because \(81 = 9^2\).

In many cases, the value is already in exponential form, like \(2^{(t/T)}\). If we took the base-2 logarithm of an exponential that already had a base of 2, then our result would just be the exponent: \(\log_2( 2^{(t/T)} ) = t/T\).

Sadly, most calculators only have [LOG] and [ln] buttons for the base-10 and base-e logarithms. There is no way to calculate a base-2 logarithm with most hand-held calculators. Though it is a bit messier, we can use the base-e logarithm to access the exponent of the previous expression: \(\ln( 2^{(t/T)} ) = (t/T)\cdot \ln(2)\), where the log of 2 is just a number (it's about 0.69).

Half-life Examples

Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. The amount remaining is multiplied by \(\frac{1}{2}\) every time a half-life elapses. We can use the formula for radioactive decay:

  • \(A\) is the amount of the substance present after time \(t\)
  • \({A}_{0}\) is the amount initially present
  • \(T\) is the half-life of the substance
  • \(t\) is the time period over which the substance is studied

It is fairly straightforward to calculate the amount remaining when given the time. If we know the amount, and need to find the time , then we either need to guess & check, or use logarithms, as shown in the next example.

Uranium-235 has a half life of 703,800,000 years. How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

After 10% decays, there will be 900 grams left. We will represent \(\frac{1}{2}\) by 0.5 in this example.

Exponential Cooling

Heat flows swiftly when a very hot object is placed in a very cold environment, but heat flows slowly when the object is only slightly warmer than the environment. When an object is at the same temperature as its environment, no heat flows at all. When objects cool down, their temperature decays exponentially. In the next example, we will use a pair of measurements to find the speed of the exponential decay, and then we will use the exponential model to predict the future temperature.

A hot iron bar is immersed in an ice-water bath to cool it down. After 10 seconds, the bar's temperature falls to 172°F, and 5 seconds later the bar's temperature is 160°F.

  • What is the "half-life" of the exponential decay?
  • What was the initial temperature, just before the bar was immersed in the ice water?
  • What will the temperature be after the bar has been cooling for 30 seconds?
  • How long must the bar cool before it is safe to handle (110°F)?

1. We know that it takes the bar 5 seconds to cool from a temperature of 172°F to 160°F. We can use that information to find the halving-time or "half-life".

\(\begin{align*}f(t)&=a\cdot(0.5)^{t/T}\\\text{ }\\160&=172\cdot(0.5)^{5/T}\end{align*}\)

The value we want to isolate is in the exponent, so we will need to use a logarithm to access it. First, we divide both sides by 172, then we take the logarithm of both sides of the equation.

\(\begin{align*}160&=172\cdot(0.5)^{5/T}\\\text{ }\\0.9302&=(0.5)^{5/T}\end{align*}\)

If we're lucky enough to have a calculator or spreadsheet that provides logarithms with arbitrary bases, we can take the logarithm with a base of 0.5.

\(\begin{align*}0.9302&=(0.5)^{5/T}\\\text{ }\\ \mathrm{log}_{0.5}(0.9302)&=\mathrm{log}_{0.5}\left((0.5)^{5/T}\right)\\\text{ }\\ \mathrm{log}_{0.5}(0.9302)&=5/T \\\text{ }\\ T\cdot \mathrm{log}_{0.5}(0.9302)&=5 \\\text{ }\\ T&=\displaystyle \frac{5}{\mathrm{log}_{0.5}(0.9302)}\\\text{ }\\ T&\approx 48\text{ seconds} \end{align*}\)

More likely, we're stuck with a calculator that only has natural and common logs. In that case, we take the logarithm of both sides of the equation using a base of either \(e\) or \(10\), and we use the properties of logarithms to isolate \(T\). We get the same result either way.

\(\begin{align*}0.9302&=(0.5)^{5/T}\\\text{ }\\ \mathrm{LOG}(0.9302)&=\mathrm{LOG}\left((0.5)^{5/T}\right)\\\text{ }\\ \mathrm{LOG}(0.9302)&=(5/T)\cdot\mathrm{LOG}(0.5) \\\text{ }\\ T\cdot\mathrm{LOG}(0.9302)&=5\cdot\mathrm{LOG}(0.5) \\\text{ }\\ T&=\displaystyle \frac{5\cdot\mathrm{LOG}(0.5)}{\mathrm{LOG}(0.9302)}\\\text{ }\\ T&\approx 48\text{ seconds} \end{align*}\)

2. To find the initial temperature, we use our half-life and our temperature of 172°F at a time of 10 seconds.

\(\begin{align*}f(t)&=a\cdot(0.5)^{t/T}\\\text{ }\\172&= a\cdot(0.5)^{10/48}\\\text{ }\\ 172&= a\cdot 0.8655\\\text{ }\\ a &\approx 199\text{°F}\end{align*}\)

3. Now that we have a complete exponential function, we can find the temperature at any point in time.

\(\begin{align*}f(t)&=a\cdot(0.5)^{t/T}\\\text{ }\\f(t)&= 199\cdot(0.5)^{t/48}\\\text{ }\\ f(30)&= 199\cdot(0.5)^{30/48}\\\text{ }\\ f(30)&\approx 129\text{°F}\end{align*}\)

4. We could guess & check different values for \(t\) until we found a temperature that was just below 110°F, but the logarithm allows us to find the answer directly.

\(\begin{align*}f(t)&=199\cdot(0.5)^{t/48}\\\text{ }\\ 110&= 199\cdot(0.5)^{t/48}\\\text{ }\\ \frac{110}{199}&= (0.5)^{t/48}\\\text{ }\\ 0.5528&= (0.5)^{t/48}\\\text{ }\\ \mathrm{LOG}(0.5528)&= \mathrm{LOG}\left((0.5)^{t/48}\right) \\\text{ }\\ \mathrm{LOG}(0.5528)&= (t/48)\cdot\mathrm{LOG}(0.5) \\\text{ }\\ \displaystyle \frac{48\cdot\mathrm{LOG}(0.5528)}{\mathrm{LOG}(0.5)}&= t \\\text{ }\\ t&\approx 41\text{ seconds}\end{align*}\)

The half-life is about 48 seconds, and the initial value was about 199°F. The bar cools to about 129°F after 30 seconds, and it needs at least 41 seconds to be safe to handle.

Exponential Heating

We typically think of exponential decay as a transition from a large value to a small one, but that is not always the case. The real key to exponential decay is that initially the rate of change is fast, and over time the rate of change slows down. For exponential cooling we have \(Temperature=RoomTemp+A\cdot(1/2)^{(t/T)}\), because the beverage starts hotter than room temperature and then cools down. When a cold beverage is placed in a hot environment, the temperature change is also exponential. With heating we have \(Temperature=RoomTemp - A\cdot(1/2)^{(t/T)}\). The only difference is that we're subtracting the exponential, so the temperature starts colder than the room and then warms up. The figure below shows examples of exponential decay downward and upward.

Continuous Growth

So far we have described exponential change using integer exponents and a decimal base, like \((1+r)^n\). When \(r\) is positive, then we have growth. When it is negative, we have decay. We have also described exponential change with bases of 2 or 1/2 and decimal exponents, like \(2^{(t/T)}\) or \((1/2)^{(t/T)}\). In some science or math contexts, it is also common to describe exponential change using the irrational number "e" for the base and a decimal exponent. This is typically less intuitive than doubling or halving, but it lends itself to other calculations involving exponential functions that are outside the scope of this course. In those cases, we describe exponential change as \(f(t)=A\cdot(e)^{(kt)}\), where \(e\) has a value of about 2.718, and \(k\) is a parameter describing the speed of the change. When \(k\gt 0\) there is growth, and when \(k\lt 0\) there is decay.

In our next example we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems. You will know to use the continuous growth or decay formula when you are asked to find an amount based on continuous compounding. In previous examples we asked that you find an amount based on quarterly or monthly compounding, and for this you used the compound interest formula.

A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?

Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. The initial investment was $1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year:

The account is worth $1,105.17 after one year.

In the following video we show another example of continuous interest.

Exponential growth or decay functions can be of the form \(f\left(t\right)=A\cdot(1+r)^{t}\) or

\(f\left(t\right)=A\cdot(B)^{t/T}\) or

\(f\left(t\right)=A\cdot(e)^{kt}\) or

\(f\left(t\right)=C\pm A\cdot(B)^{t/T}\).

We can relate \(r\) and \(T\) using the approximation \(70/R\approx T\) (where \(R\) is just \(r\) without the percent sign).

We can also relate \(r\) and \(T\) by using logarithms, which convert a number into exponential form and access the resulting exponent.

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4.6: Exponential and Logarithmic Models

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While we have explored some basic applications of exponential and logarithmic functions, in this section we explore some important applications in more depth.

Radioactive Decay

In an earlier section, we discussed radioactive decay – the idea that radioactive isotopes change over time. One of the common terms associated with radioactive decay is half-life.

Definition: Half Life

The half-life of a radioactive isotope is the time it takes for half the substance to decay.

Given the basic exponential growth/decay equation \(h(t)=ab^{t}\), half-life can be found by solving for when half the original amount remains; by solving \(\dfrac{1}{2} a=a(b)^{t}\), or more simply \(\dfrac{1}{2} =b^{t}\). Notice how the initial amount is irrelevant when solving for half-life.

Example \(\PageIndex{1}\)

Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210?

We were not given a starting quantity, so we could either make up a value or use an unknown constant to represent the starting amount. To show that starting quantity does not affect the result, let us denote the initial quantity by the constant a . Then the decay of Bismuth-210 can be described by the equation \(Q(d)=a(0.87)^{d}\).

To find the half-life, we want to determine when the remaining quantity is half the original: \(\dfrac{1}{2} a\). Solving,

\[\dfrac{1}{2} a=a(0.87)^{d}\nonumber\] Divide by \(a\),

\[\dfrac{1}{2} =0.87^{d}\nonumber\] Take the log of both sides

\[\log \left(\dfrac{1}{2} \right)=\log \left(0.87^{d} \right)\nonumber\] Use the exponent property of logs

\[\log \left(\dfrac{1}{2} \right)=d\log \left(0.87\right)\nonumber\] Divide to solve for \(d\)

\[d=\dfrac{\log \left(\dfrac{1}{2} \right)}{\log \left(0.87\right)} \approx 4.977\text{ days}\nonumber \]

This tells us that the half-life of Bismuth-210 is approximately 5 days.

Example \(\PageIndex{2}\)

Cesium-137 has a half-life of about 30 years. If you begin with 200 mg of cesium-137, how much will remain after 30 years? 60 years? 90 years?

Since the half-life is 30 years, after 30 years, half the original amount, 100 mg, will remain.

After 60 years, another 30 years have passed, so during that second 30 years, another half of the substance will decay, leaving 50 mg.

After 90 years, another 30 years have passed, so another half of the substance will decay, leaving 25 mg.

Example \(\PageIndex{3}\)

Cesium-137 has a half-life of about 30 years. Find the annual decay rate.

Since we are looking for an annual decay rate, we will use an equation of the form \(Q(t)=a(1+r)^{t}\). We know that after 30 years, half the original amount will remain. Using this information

\[\dfrac{1}{2} a=a(1+r)^{30}\nonumber\] Dividing by \(a\)

\[\dfrac{1}{2} =(1+r)^{30}\nonumber\] Taking the 30\({}^{th}\) root of both sides

\[\sqrt[{30}]{\dfrac{1}{2} } =1+r\nonumber\] Subtracting one from both sides,

\[r=\sqrt[{30}]{\dfrac{1}{2} } -1\approx -0.02284\nonumber\]

This tells us cesium-137 is decaying at an annual rate of 2.284% per year.

Exercise \(\PageIndex{1}\)

Chlorine-36 is eliminated from the body with a biological half-life of 10 days (www.ead.anl.gov/pub/doc/chlorine.pdf). Find the daily decay rate.

\(r = \sqrt[10]{\dfrac{1}{2}} - 1 \approx -0.067\) or 6.7% is the daily rate of decay.

Example \(\PageIndex{4}\)

Carbon-14 is a radioactive isotope that is present in organic materials, and is commonly used for dating historical artifacts. Carbon-14 has a half-life of 5730 years. If a bone fragment is found that contains 20% of its original carbon-14, how old is the bone?

To find how old the bone is, we first will need to find an equation for the decay of the carbon-14. We could either use a continuous or annual decay formula, but opt to use the continuous decay formula since it is more common in scientific texts. The half life tells us that after 5730 years, half the original substance remains. Solving for the rate,

\[\dfrac{1}{2} a=ae^{r5730}\nonumber\] Dividing by \(a\)

\[\dfrac{1}{2} =e^{r5730}\nonumber\] Taking the natural log of both sides

\[\ln \left(\dfrac{1}{2} \right)=\ln \left(e^{r5730} \right)\nonumber\] Use the inverse property of logs on the right side

\[\ln \left(\dfrac{1}{2} \right)=5730r\nonumber\] Divide by 5730

\[r=\dfrac{\ln \left(\dfrac{1}{2} \right)}{5730} \approx -0.000121\nonumber\]

Now we know the decay will follow the equation \(Q(t)=ae^{-0.000121t}\). To find how old the bone fragment is that contains 20% of the original amount, we solve for \(t\) so that \(Q(t) = 0.20a\).

\[0.20a=ae^{-0.000121t}\nonumber\] \[0.20=e^{-0.000121t}\nonumber\] \[\ln (0.20)=\ln \left(e^{-0.000121t} \right)\nonumber\] \[\ln (0.20)=-0.000121t\nonumber\] \[t=\dfrac{\ln (0.20)}{-0.000121} \approx 13301\text{ years}\nonumber\]

The bone fragment is about 13,300 years old.

Exercise \(\PageIndex{2}\)

In Example 2, we learned that Cesium-137 has a half-life of about 30 years. If you begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

Less than 230 years, 229.3157 to be exact.

Doubling Time

For decaying quantities, we asked how long it takes for half the substance to decay. For growing quantities we might ask how long it takes for the quantity to double.

Definition: Doubling Time

The doubling time of a growing quantity is the time it takes for the quantity to double.

Given the basic exponential growth equation \(h(t)=ab^{t}\), doubling time can be found by solving for when the original quantity has doubled; by solving \(2a=a(b)^{x}\), or more simply \(2=b^{x}\). Like with decay, the initial amount is irrelevant when solving for doubling time.

Example \(\PageIndex{5}\)

Cancer cells sometimes increase exponentially. If a cancerous growth contained 300 cells last month and 360 cells this month, how long will it take for the number of cancer cells to double?

Defining \(t\) to be time in months, with \(t = 0\) corresponding to this month, we are given two pieces of data: this month, (0, 360), and last month, (-1, 300).

From this data, we can find an equation for the growth. Using the form \(C(t)=ab^{t}\), we know immediately a = 360, giving \(C(t)=360b^{t}\). Substituting in (-1, 300), \[\begin{array}{l} {300=360b^{-1} } \\ {300=\dfrac{360}{b} } \\ {b=\dfrac{360}{300} =1.2} \end{array}\nonumber\]

This gives us the equation \(C(t)=360(1.2)^{t}\)

To find the doubling time, we look for the time when we will have twice the original amount, so when \(C(t) = 2a\).

\[2a=a(1.2)^{t}\nonumber\] \[2=(1.2)^{t}\nonumber\] \[\log \left(2\right)=\log \left(1.2^{t} \right)\nonumber\] \[\log \left(2\right)=t\log \left(1.2\right)\nonumber\] \[t=\dfrac{\log \left(2\right)}{\log \left(1.2\right)} \approx 3.802\nonumber\] months for the number of cancer cells to double.

Example \(\PageIndex{6}\)

Use of a new social networking website has been growing exponentially, with the number of new members doubling every 5 months. If the site currently has 120,000 users and this trend continues, how many users will the site have in 1 year?

We can use the doubling time to find a function that models the number of site users, and then use that equation to answer the question. While we could use an arbitrary a as we have before for the initial amount, in this case, we know the initial amount was 120,000.

If we use a continuous growth equation, it would look like \(N(t)=120e^{rt}\), measured in thousands of users after t months. Based on the doubling time, there would be 240 thousand users after 5 months. This allows us to solve for the continuous growth rate:

\[240=120e^{r5}\nonumber\] \[2=e^{r5}\nonumber\] \[\ln 2=5r\nonumber\] \[r=\dfrac{\ln 2}{5} \approx 0.1386\nonumber\]

Now that we have an equation, \(N(t)=120e^{0.1386t}\), we can predict the number of users after 12 months:

\[N(12) =120e^{0.1386(12)} =633.140\text{ thousand users}\nonumber\].

So after 1 year, we would expect the site to have around 633,140 users.

Exercise \(\PageIndex{3}\)

If tuition at a college is increasing by 6.6% each year, how many years will it take for tuition to double?

Solving \(a (1 + 0.066)^t = 2a\), it will take \(t = \dfrac{log(2)}{log(1.066)} \approx 10.845\) years, or approximately 11 years, for tuition to double.

Newton’s Law of Cooling

When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off towards the surrounding air temperature. This "leveling off" will correspond to a horizontal asymptote in the graph of the temperature function. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function.

Definition: Newton’s Law of Cooling

The temperature of an object, \(T\), in surrounding air with temperature \(T_{s}\) will behave according to the formula

\[T(t)=ae^{kt} +T_{s}\]

  • \(t\) is time
  • \(a\) is a constant determined by the initial temperature of the object
  • \(k\) is a constant, the continuous rate of cooling of the object

While an equation of the form \(T(t)=ab^{t} +T_{s}\) could be used, the continuous growth form is more common.

Example \(\PageIndex{7}\)

A cheesecake is taken out of the oven with an ideal internal temperature of 165 degrees Fahrenheit, and is placed into a 35 degree refrigerator. After 10 minutes, the cheesecake has cooled to 150 degrees. If you must wait until the cheesecake has cooled to 70 degrees before you eat it, how long will you have to wait?

Since the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially towards 35, following the equation

\[T(t)=ae^{kt} +35\nonumber\]

We know the initial temperature was 165, so \(T(0)=165\). Substituting in these values,

\[\begin{array}{l} {165=ae^{k0} +35} \\ {165=a+35} \\ {a=130} \end{array}\nonumber\]

We were given another pair of data, \(T(10)=150\), which we can use to solve for \(k\)

\[150=130e^{k10} +35\nonumber\] \[\begin{array}{l} {115=130e^{k10} } \\ {\dfrac{115}{130} =e^{10k} } \\ {\ln \left(\dfrac{115}{130} \right)=10k} \\ {k=\dfrac{\ln \left(\dfrac{115}{130} \right)}{10} =-0.0123} \end{array}\nonumber\]

Together this gives us the equation for cooling: \[T(t)=130e^{-0.0123t} +35\nonumber\]

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

\[70=130e^{-0.0123t} +35\nonumber\] \[35=130e^{-0.0123t}\nonumber\] \[\dfrac{35}{130} =e^{-0.0123t}\nonumber\] \[\ln \left(\dfrac{35}{130} \right)=-0.0123t\nonumber\] \[t=\dfrac{\ln \left(\dfrac{35}{130} \right)}{-0.0123} \approx 106.68\nonumber\]

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool. Of course, if you like your cheesecake served chilled, you’d have to wait a bit longer.

Exercise \(\PageIndex{4}\)

A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?

\(T(t) = ae^{kt} + 70\). Substituting (0, 40), we find \(a = -30\). Substituting (1, 45), we solve \[45 = -30 e^{k(1)} + 70\nonumber\] to get \[k = ln(\dfrac{25}{30}) = -0.1823\nonumber\]

Solving \(60 = -30e^{-0.1823t} + 70\) gives

\[t = \dfrac{ln(1/3)}{-0.1823} = 6.026\text{ hours}\nonumber \]

  • Logarithmic Scales

For quantities that vary greatly in magnitude, a standard scale of measurement is not always effective, and utilizing logarithms can make the values more manageable. For example, if the average distances from the sun to the major bodies in our solar system are listed, you see they vary greatly.

Placed on a linear scale – one with equally spaced values – these values get bunched up.

A number line labeled distance from 0 to 4500 with equally spaced tick marks every 250. There are arrows pointing to locations for each planet. There are four arrows between 0 and 250 for Mercury Venus Earth and Mars, then an arrow at 779 for Jupiter, at 1430 for Saturn, at 2880 for Uranus, and at 4500 for Neptune.

0 500 1000 1500 2000 2500 3000 3500 4000 4500

However, computing the logarithm of each value and plotting these new values on a number line results in a more manageable graph, and makes the relative distances more apparent.(It is interesting to note the large gap between Mars and Jupiter on the log number line. The asteroid belt is located there, which scientists believe is a planet that never formed because of the effects of the gravity of Jupiter.)

A numberline labeled log of distance, from 1.5 to 4.  At 2 there's an arrow noting 10 squared equals 100. At 3 there's an arrow noting 10 cubed equals 1000.  There is an arrow of Mercury at 1.76, for Venus at 2.03, for Earth at 2.18, and additional arrow for the other planets at the log of distance values from the table above.

Sometimes, as shown above, the scale on a logarithmic number line will show the log values, but more commonly the original values are listed as powers of 10, as shown below.

A numberline with equally spaced ticks labeled 10 to the negative 2, 10 to the negative 1, 10 to the zero, 10 to the 1, 10 to the 2, and so on up to 10 to the 7.  There are 5 points labeled: P at 10 to the negative 1.5, A at 10 to the 1, B at 10 to the 2, C at 10 to the 5, and D at 10 to the 6.

Example \(\PageIndex{8}\)

Estimate the value of point \(P\) on the log scale above

The point \(P\) appears to be half way between -2 and -1 in log value, so if \(V\) is the value of this point,

\[\log (V)\approx -1.5\nonumber\] Rewriting in exponential form, \[V\approx 10^{-1.5} =0.0316\nonumber\]

Example \(\PageIndex{9}\)

Place the number 6000 on a logarithmic scale.

Since \(\log (6000)\approx 3.8\), this point would belong on the log scale about here:

A numberline with equally spaced ticks labeled 10 to the negative 2, 10 to the negative 1, 10 to the zero, 10 to the 1, 10 to the 2, and so on up to 10 to the 7.  There is a point labeled 6000 placed partway between 10 to the 3.5 and 10 to the 4.

Exercise \(\PageIndex{5}\)

Plot the data in the table below on a logarithmic scale (From http://www.epd.gov.hk/epd/noise_educ...1/intro_5.html , retrieved Oct 2, 2010).

A numberline with equally spaced ticks labeled 10 to the 1, 10 to the 2, and so on up to 10 to the 10.  There are arrows for Softest sound around 10 to the 1.3, Broadcast room around 10 to the 2.3, Soft whisper at 10 to the 3.3, Conversation at 10 to the 4.3, Train at 10 to the 5.3, Symphony at 10 to the 6.3, and Space shuttle at 10 to the 9.3.

Notice that on the log scale above Example 8, the visual distance on the scale between points \(A\) and \(B\) and between \(C\) and \(D\) is the same. When looking at the values these points correspond to, notice \(B\) is ten times the value of \(A\), and \(D\) is ten times the value of \(C\). A visual \(linear\) difference between points corresponds to a relative (ratio) change between the corresponding values.

Logarithms are useful for showing these relative changes. For example, comparing $1,000,000 to $10,000, the first is 100 times larger than the second.

\[\dfrac{1,000,000}{10,000} = 100 = 10^2\nonumber\]

Likewise, comparing $1000 to $10, the first is 100 times larger than the second.

\[\dfrac{1,000}{10} = 100 = 10^2\nonumber\]

When one quantity is roughly ten times larger than another, we say it is one order of magnitude larger. In both cases described above, the first number was two orders of magnitude larger than the second.

Notice that the order of magnitude can be found as the common logarithm of the ratio of the quantities. On the log scale above, B is one order of magnitude larger than \(A\), and \(D\) is one order of magnitude larger than \(C\).

Definition: Orders of magnitude

Given two values \(A\) and \(B\), to determine how many orders of magnitude \(A\) is greater than \(B\),

Difference in orders of magnitude = log(\(\dfrac{A}{B})\)

Example \(\PageIndex{10}\)

On the log scale above Example 8, how many orders of magnitude larger is \(C\) than \(B\)?

The value \(B\) corresponds to \(10^2 = 100\)

The value \(C\) corresponds to \(10^5 = 100,000\)

The relative change is \(\dfrac{100,000}{100} = 1000 = \dfrac{10^5}{10^2} = 10^3\). The log of this value is 3.

\(C\) is three orders of magnitude greater than \(B\), which can be seen on the log scale by the visual difference between the points on the scale.

Exercise \(\PageIndex{6}\)

Using the table from Try it Now #5, what is the difference of order of magnitude between the softest sound a human can hear and the launching of the space shuttle?

\(\dfrac{2 \times 10^9}{2 \times 10^1} = 10^8\). The sound pressure in \(\mu\)Pa created by launching the space shuttle is 8 orders of magnitude greater than the sound pressure in \(\mu\)Pa created by the softest sound a human ear can hear.

Earthquakes

An example of a logarithmic scale is the Moment Magnitude Scale (MMS) used for earthquakes. This scale is commonly and mistakenly called the Richter Scale, which was a very similar scale succeeded by the MMS.

Moment Magnitude Scale

For an earthquake with seismic moment \(S\), a measurement of earth movement, the MMS value, or magnitude of the earthquake, is

\[M = \dfrac{2}{3} log(\dfrac{S}{S_0})\]

Where \(S_0 = 10^{16}\) is a baseline measure for the seismic moment.

Example \(\PageIndex{11}\)

If one earthquake has a MMS magnitude of 6.0, and another has a magnitude of 8.0, how much more powerful (in terms of earth movement) is the second earthquake?

Since the first earthquake has magnitude 6.0, we can find the amount of earth movement for that quake, which we'll denote \(S_1\). The value of \(S_0\) is not particularity relevant, so we will not replace it with its value.

\[6.0 = \dfrac{2}{3} log (\dfrac{S_1}{S_0})\nonumber\] \[6.0 (\dfrac{3}{2} = log (\dfrac{S_1}{S_0})\nonumber\] \[9 = log(\dfrac{S_1}{S_0})\nonumber\] \[\dfrac{S_1}{S_0} = 10^9\nonumber\] \[S_1 = 10^9 S_0\nonumber\]

This tells us the first earthquake has about \(10^9\) times more earth movement than the baseline measure.

Doing the same with the second earthquake, \(S_2\), with a magnitude of 8.0,

\[8.0 = \dfrac{2}{3} log (\dfrac{S_2}{S_0})\nonumber\] \[S_2 = 10^{12} S_0\nonumber\]

Comparing the earth movement of the second earthquake to the first,

\[\dfrac{S_2}{S_1} = \dfrac{10^{12} S_0} {10^9 S_0} = 10^3 = 1000\nonumber\]

The second value's earth movement is 1000 times as large as the first earthquake.

Example \(\PageIndex{12}\)

One earthquake has magnitude of 3.0. If a second earthquake has twice as much earth movement as the first earthquake, find the magnitude of the second quake.

Since the first quake has magnitude 3.0,

\[3.0 = \dfrac{2}{3} log (\dfrac{S}{S_0})\nonumber\]

Solving for \(S\),

\[3.0 \dfrac{3}{2} = log (\dfrac{S}{S_0})\nonumber\] \[4.5 = log (\dfrac{S}{S_0})\nonumber\] \[10^{4.5} = \dfrac{S}{S_0}\nonumber\] \[S = 10^{4.5} S_0\nonumber\]

Since the second earthquake has twice as much earth movement, for the second quake,

\[S = 2 \cdot 10^{4.5} S_0\nonumber\]

Finding the magnitude,

\[M = \dfrac{2}{3} log (\dfrac{2 \cdot 10^{4.5} S_0}{S_0})\nonumber\] \[M = \dfrac{2}{3} log (2 \cdot 10^{4.5}) \approx 3.201\nonumber\]

The second earthquake with twice as much earth movement will have a magnitude of about 3.2.

In fact, using log properties, we could show that whenever the earth movement doubles, the magnitude will increase by about 0.201:

\[M = \dfrac{2}{3} log (\dfrac{2S}{S_0}) = \dfrac{2}{3} log (2 \cdot \dfrac{S}{S_0})\nonumber\] \[M = \dfrac{2}{3} (log(2) + log(\dfrac{S}{S_0}))\nonumber\] \[M = \dfrac{2}{3} log (2) + \dfrac{2}{3} log (\dfrac{S}{S_0})\nonumber\] \[M = 0.201 + \dfrac{2}{3} log (\dfrac{S}{S_0})\nonumber\]

This illustrates the most important feature of a log scale: that \(multiplying\) the quantity being considered will \(add\) to the scale value, and vice versa.

Important Topics of this Section

  • Radioactive decay
  • Doubling time
  • Newton’s law of cooling
  • Orders of Magnitude
  • Moment Magnitude scale

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College Algebra

Unit 1: linear equations and inequalities, unit 2: graphs and forms of linear equations, unit 3: functions, unit 4: quadratics: multiplying and factoring, unit 5: quadratic functions and equations, unit 6: complex numbers, unit 7: exponents and radicals, unit 8: rational expressions and equations, unit 9: relating algebra and geometry, unit 10: polynomial arithmetic, unit 11: advanced function types, unit 12: transformations of functions, unit 13: rational exponents and radicals, unit 14: logarithms.

Lesson Radioactive decay problems

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How to Calculate Half Life

Last Updated: May 26, 2023 References

Understanding Half-Life

Learning the half-life equation, calculating from a graph, using a calculator, example problems, calculator, practice problems, and answers, expert q&a.

This article was co-authored by Meredith Juncker, PhD and by wikiHow staff writer, Hannah Madden . Meredith Juncker is a PhD candidate in Biochemistry and Molecular Biology at Louisiana State University Health Sciences Center. Her studies are focused on proteins and neurodegenerative diseases. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been viewed 1,152,613 times.

The half-life of a substance undergoing decay is the time it takes for the amount of the substance to decrease by half. It was originally used to describe the decay of radioactive elements like uranium or plutonium, but it can be used for any substance which undergoes decay along a set, or exponential, rate. You can calculate the half-life of any substance, given the rate of decay, which is the initial quantity of the substance and the quantity remaining after a measured period of time. [1] X Research source

Step 1 What is half-life?

  • Elements like uranium and plutonium are most often studied with half-life in mind.

Step 2 Does temperature or concentration affect the half-life?

  • Therefore, you can calculate the half-life for a particular element and know for certain how quickly it will break down no matter what.

Step 3 Can half-life be used in carbon dating?

  • Technically, there are 2 types of carbon: carbon-14, which decays, and carbon-12, which stays constant.

Step 1 Understand exponential decay.

  • Simply replacing the variable doesn't tell us everything, though. We still have to account for the actual half-life, which is, for our purposes, a constant.

t_{{1/2}}

  • On half-life graphs, the x-axis will usually show the timeline, while the y-axis usually shows the rate of decay.

Step 2 Go down half the original count rate and mark it on the graph.

  • For example, if the starting point is 1,640, divide 1,640 / 2 to get 820.
  • If you are working with a semi log  plot, meaning the count rate is not evenly spaced, you’ll have to take the logarithm of any number from the vertical axis. [11] X Research source

Step 3 Draw a vertical line down from the curve.

  • If you know the half-life but you don’t know the initial quantity, you can input the half-life, the quantity that remains, and the time that has passed. As long as you know 3 of the 4 values, you’ll be able to use a half-life calculator.

Step 2 Calculate the decay constant with a half-life calculator.

  • If you don’t know the half-life but you do know the decay constant and the mean lifetime, you can input those instead. Just like the initial equation, you only need to know 2 of the 3 values to get the third one.

Step 3 Plot your half-life equation on a graphing calculator.

  • This is a helpful visual, and it can be useful if you don’t want to do all of the equation work.

Step 1 Problem 1.

  • Check to see if the solution makes sense. Since 112 g is less than half of 300 g, at least one half-life must have elapsed. Our answer checks out.

Step 2 Problem 2.

  • Substitute and evaluate.

{\begin{aligned}t&=(70{{\rm {\ years}}})\log _{{1/2}}\left({\frac  {0.1{{\rm {\ kg}}}}{20{{\rm {\ kg}}}}}\right)\\&\approx 535{{\rm {\ years}}}\end{aligned}}

  • Remember to check your solution intuitively to see if it makes sense.

Step 3 Problem 3.

  • For this particular equation, the actual length of the half-life did not play a role.

Step 4 Problem 4.

Video . By using this service, some information may be shared with YouTube.

N(t)

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Calculate Wavelength

  • ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Half-lives_and_Pharmacokinetics
  • ↑ https://chem.libretexts.org/Courses/Furman_University/CHM101%3A_Chemistry_and_Global_Awareness_(Gordon)/05%3A_Basics_of_Nuclear_Science/5.07%3A_Calculating_Half-Life
  • ↑ https://atomic.lindahall.org/what-is-meant-by-half-life.html
  • ↑ http://faculty.bard.edu/belk/math213/ExponentialDecay.pdf
  • ↑ https://www.ausetute.com.au/halflife.html
  • ↑ https://socratic.org/chemistry/nuclear-chemistry/nuclear-half-life-calculations
  • ↑ https://www.khanacademy.org/test-prep/mcat/physical-processes/atomic-nucleus/a/decay-graphs-and-half-lives-article
  • ↑ https://www.gcsescience.com/prad17-measuring-half-life.htm
  • ↑ https://www.calculator.net/half-life-calculator.html
  • ↑ https://www.youtube.com/watch?v=5mKrIv1lo1E&feature=youtu.be&t=163
  • ↑ https://www.chemteam.info/Radioactivity/Radioactivity-Half-Life-probs1-10.html

About This Article

Meredith Juncker, PhD

To find the half life of a substance, or the time it takes for a substance to decrease by half, you’ll be using a variation of the exponential decay formula. Plug in ½ for a, use the time for x, and multiply the left side by the initial quantity of the substance. Rearrange the equation so that you’re solving for what the problem asks for, whether that’s half life, mass, or another value. Plug in the values you have and solve, writing the answer in seconds, days, or years. To see the half life equation and look at examples, read on! Did this summary help you? Yes No

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  • Chemistry Concept Questions and Answers
  • Half Life Questions

Half-Life Questions

Half-Life or previously known as the Half-Life Period is one of the common terminologies used in Science to describe the radioactive decay of a particular sample or element within a certain period of time.

However, this concept is also widely used to describe various types of decay processes, especially exponential and non-exponential decay. Apart from science, the term is used in medical sciences to represent the biological half-life of certain chemicals in the human body or in drugs.

Half-Life Chemistry Questions with Solutions

Q1. An isotope of caesium (Cs-137) has a half-life of 30 years. If 1.0g of Cs-137 disintegrates over a period of 90 years, how many grams of Cs-137 would remain?

b.) 0.125 g

c.) 0.00125 g

Correct Answer- (b.) 0.125 g

Q2. Selenium-83 has a half-life of 25.0 minutes. How many minutes would it take for a 10.0 mg sample to decay and only have 1.25 mg of it remain?

a.) 75 minutes

b.) 75 days

c.) 75 seconds

d.) 75 hours

Correct Answer- (a.) 75 minutes

Q3. How long does it take a 100.00g sample of As-81, with a half-life of 33 seconds, to decay to 6.25g?

a.) 122 seconds

b.) 101 seconds

c.) 132 seconds

d.) 22 seconds

Correct Answer – (c.) 132 seconds

Q4. What is the half-life of a radioactive isotope if a 500.0g sample decays to 62.5g in 24.3 hours?

a.) 8.1 hours

b.) 6.1 hours

c.) 5 hours

d.) 24 hours

Correct Answer- (a.) 8.1 hours

Q5. What is the half-life of Polonium-214 if, after 820 seconds, a 1.0g sample decays to 0.03125g?

a.) 164 minutes

b.) 164 seconds

c.) 64 seconds

d.) 160 minutes

Correct Answer- (b.) 164 seconds

Q6. The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes have elapsed?

To begin, we’ll count the number of half-lives that have passed. This can be obtained by doing the following:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

Number of half-lives

n = 7.2/2.4 = 3

Thus, three half-lives have passed.

Finally, we will calculate the remaining amount. This can be obtained by doing the following:

N 0 (original amount) = 100 g

(n) = number of half-lives

Amount remaining (N) =?

N = 100 / 2 3

N = 100 / 8

As a result, the amount of Zn-71 remaining after 7.2 minutes is 12.5 g.

Q7. Pd-100 has a half-life of 3.6 days. If one had 6.02 x 10 23 atoms at the start, how many atoms would be present after 20.0 days?

Half-life = 3.6 days

Initial atoms = 6.02 ×10 23 atoms

Time = 20days

To calculate the atoms present after 20 days, we use the formula below.

Thus, the number of atoms available is 1.28 × 10 22 atoms.

Q8. Os-182 has a half-life of 21.5 hours. How many grams of a 10.0 gram sample would have decayed after exactly three half-lives?

Answer. The amount of the radioactive substance that will remain after 3- half- lives=(½) 3 × a,

where a = initial concentration of the radioactive element.

So, amount of the radioactive substance that remains aftet 3- half-lives=( ½)³x10 = 10/8= 1.25 g.

Therefore, the number of grams of the radioactive substance that decayed in 3 half-lives = (10 – 1.25) g

Q9. After 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

Answer. The remaining decimal fraction is:

2.00 mg / 128.0 mg = 0.015625

The half-lives that must have expired to get to 0.015625?

(½) n = 0.015625

n log 0.5 = 0.015625

n = log 0.5 / 0.015625 n = 6

Calculation of the half-life:

24 days divided by 6 half-lives equals 4.00 days

Q10. A radioactive isotope decayed to 17/32 of its original mass after 60 minutes. Find the half-life of this radioisotope.

Answer. The amount that remains

17/32 = 0.53125

(1/2) n = 0.53125

n log 0.5 = log 0.53125

n = 0.91254

Half-lives that have elapsed are therefore, n = 0.9125

60 minutes divided by 0.91254 equals 65.75 minutes.

Therefore, n = 66 minutes

Q11. How long will it take for a 40 gram sample of I-131 (half-life = 8.040 days) to decay to 1/100 of its original mass?

Answer. (1/2) n = 0.01

n log 0.5 = log 0.01

6.64 x 8.040 days = 53.4 days

Therefore, it will take 53.4 days to decay to 1/100 of its original mass.

Q12. At time zero, there are 10.0 grams of W-187. If the half-life is 23.9 hours, how much will be present at the end of one day? Two days? Seven days?

24.0 hr / 23.9 hr/half-life = 1.0042 half-lives

One day = one half-life; (1/2) 1.0042 = 0.4985465 remaining = 4.98 g

Two days = two half-lives; (1/2) 2.0084 = 0.2485486 remaining = 2.48 g

Seven days = 7 half-lives; (1/2) 7.0234 = 0.0076549 remaining = 0.0765 g

Q13. 100.0 grams of an isotope with a half-life of 36.0 hours is present at time zero. How much time will have elapsed when 5.00 grams remains?

The afraction amount remaining will be-

5.00 / 100.0 = 0.05

(1/2) n = 0.05

n log 0.5 = log 0.05

n = 4.32 half-lives

36.0 hours x 4.32 = 155.6 hours

Q14. How much time will be required for a sample of H-3 to lose 75% of its radioactivity? The half-life of tritium is 12.26 years.

If you lose 75%, then 25% remains.

(1/2) n = 0.25

n = 2 (Since, (1/2) 2 = 1/4 and 1/4 = 0.25)

12.26 x 2 = 24.52 years

Therefore, 24.52 years of time will be required for a sample of H-3 to lose 75% of its radioactivity

Q15. The half-life for the radioactive decay of 14 C is 5730 years. An archaeological artifact containing wood had only 80% of the 14 C found in a living tree. Estimate the age of the sample.

Answer. Decay constant, k = 0.693/t 1/2 = 0.693/5730 years = 1/209 × 10 –4 /year

= 1846 years (approx)

Practise Questions on Half-Life

Q1. A newly prepared radioactive nuclide has a decay constant λ of 10 –6 s –1 . What is the approximate half-life of the nuclide?

d.) 1 month

Q2. If the decay constant of a radioactive nuclide is 6.93 x 10 –3 sec –1 , its half-life in minutes is:

Q3. A first-order reaction takes 40 min for 30% decomposition. Calculate t 1/2 .

Q4. What will be the time for 50% completion of a first-order reaction if it takes 72 min for 75% completion?

Q5. How much time will it take for 90% completion of a reaction if 80% of a first-order reaction was completed in 70 min?

Click the PDF to check the answers for Practice Questions. Download PDF

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Half-Life Formula: Components and Applications

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Man in lab coat measuring a decayed skull in a white laboratory

In nuclear physics, the concept of half-life plays a crucial role in understanding the decay of radioactive substances. Scientists use the half-life formula in other disciplines to predict the rate of decay, as well as measure the age of ancient artifacts through carbon dating.

What Is Half-life?

Components of the half-life formula, applications of the half-life formula.

In the context of radioactive decay and nuclear physics, half-life describes the time it takes for half of a quantity of a substance undergoing decay to go through transformation. In simpler terms, half-life is the duration it takes for a radioactive substance to lose half of its initial radioactivity.

For example, if you start with a certain amount of a radioactive substance, after one half-life, half of that substance will have decayed, and you will have half of the original amount. After two half-lives, three-quarters will have decayed, and so on.

Radioactive Decay and Isotopes

Half-life is a characteristic property of each radioactive isotope , and it plays a crucial role in understanding the stability and decay of atomic nuclei. You can express the concept mathematically through an exponential decay model, where the rate of decay is proportional to the remaining quantity of the substance.

The half-life of a radioactive isotope — denoted by T 1/2 — varies widely depending on the specific isotope . Each has its own unique half-life. Some isotopes have very short half-lives, measured in seconds or minutes, while others have half-lives that extend over thousands or millions of years.

The concept of half-life is not limited to radioactive decay; other fields like medicine, chemistry and environmental science also measure half-life.

The half-life formula is:

Here are the formula's different components:

  • N(t) represents the remaining quantity of the radioactive substance at time t .
  • N is the initial quantity of the substance at time t = 0 . The initial quantity sets the starting point.
  • e is the base of the natural logarithm.
  • λ is the decay constant, a measure of the rate of decay for the radioactive isotope.

In the formula, e –λt is the core exponential decay factor, governing the decrease in quantity over time. As time ( t ) increases, this factor approaches 0, indicating an exponentially decaying quantity due to radioactive decay.

You can find a half-life calculator online to simplify the process of solving half-life problems.

Here are a few uses for the half-life formula.

  • Radioactive decay : Scientists use the half-life formula to describe the decay process of radioactive isotopes. It helps determine the rate of decay and predict how much of a substance will remain after a certain period.
  • Carbon dating : Carbon dating relies on the half-life of carbon-14 ( 14 C) to estimate the age of organic materials. By measuring the ratio of carbon-14 to carbon-12 in a sample, scientists can calculate how many half-lives have elapsed since the organism died.
  • Archaeology and geology : Half-life calculations are essential in dating ancient artifacts and geological samples. For example, scientists can use the decay of uranium to lead to determine the age of rocks.
  • Medical imaging : Radioactive substances used in medical imaging have known half-lives. Understanding the half-life allows medical professionals to determine the appropriate dosage and timing for imaging procedures.

This article was created in conjunction with AI technology, then was fact-checked and edited by a HowStuffWorks editor.

Please copy/paste the following text to properly cite this HowStuffWorks.com article:

For more audio journalism and storytelling, download New York Times Audio , a new iOS app available for news subscribers.

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COMMENTS

  1. Solving half-life problems with exponential decay

    Finding HALF LIFE (KristaKingMath) Share. Watch on. Growth and decay problems are another common application of derivatives. We actually don't need to use derivatives in order to solve these problems, but derivatives are used to build the basic growth and decay formulas, which is why we study these applications in this part of calculus.

  2. Exponential Growth and Decay

    The equation is y=2e3x y = 2 e 3 x. A graph showing exponential decay. The equation is y=3e−2x y = 3 e − 2 x. Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude.

  3. Constructing exponential models: half life

    8 years ago When we say that a mass is "halved", we don't mean that, after 5730 years, half suddenly disappears. We're saying that atoms of carbon-14 (let's say there are 1000) are disappearing at a certain rate (we don't know the exact rate) which is decelerating (so we lose less every 5730 years).

  4. Solving Half-Life Problems

    In this video we go over the equation for solving half-life problems. Then we do some examples where we solve for the amount of the substance you have left ...

  5. How to Find Half-Life

    Half-life: The half-life of a radioactive substance is the amount of time it takes for exactly half of the atoms in a substance to decay to a stable isotope. Half-life Formula: The...

  6. 6.7 Exponential and Logarithmic Models

    We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

  7. 4.3: Special Cases- Doubling Time and Half-Life

    The half-life is given in minutes and we want to know how much is left in two hours. Convert hours to minutes when using the model: two hours = 120 minutes. P0 = 1000 and H = 58 minutes, so the half-life model for this problem is: P(t) = 1000(1 2) t 58. With t = 120 minutes, P(120) = 1000(1 2)120 58 = 238.33.

  8. College Algebra

    0:00 / 7:56 College Algebra - Exponential Decay Problem, Half Life Math Widget 698 subscribers Subscribe 3 800 views 7 years ago College Algebra This problem looks at how half...

  9. How to Solve Half-Life Problems : Fun With Math

    Subscribe Now:http://www.youtube.com/subscription_center?add_user=ehoweducationWatch More:http://www.youtube.com/ehoweducationSolving half-life problems is s...

  10. College Algebra Tutorial 47

    College Algebra Tutorial 47: Exponential Growth and Decay. WTAMU> Virtual Math Lab > College Algebra. Learning Objectives. After completing this tutorial, you should be able to: Solve exponential growth problems. Solve exponential decay problems. Introduction. In this tutorial I will step you through how to solve problems that deal in ...

  11. PDF Half-life Problems

    1 6 = _____ ( ) 2 75 = 100 (1 2) 6 HINT: When you use rational exponents in your calculator, put ( ) around them! = _____________ After 75 hours, about __________ mg of technetium-99 remains. Use a similar format in order to find the equations and solutions of the 4 remaining problems. 2) Arsenic-74 is used to locate brain tumors.

  12. What is exponential decay? How does it work?

    To solve exponential decay word problems, you may be plugging one value into the exponential-decay equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the half-life information to find the decay constant (probably by ... College Math. College Pre-Algebra; Introductory Algebra;

  13. Half-life (qualitative) (practice)

    Lesson 4: How do we determine the age of fossils? Half-life and carbon dating. Half-life (qualitative) Half life (intermediate) Worked example: Half-life. Worked example: Fraction of undecayed nuclei. Activity and Mean life. Potassium-argon (K-Ar) dating. K-Ar dating calculation.

  14. Half Life Formula

    Solution: Given decay constant λ = 0.84 The half life formula can be used to find the half life of the substance. t1 2 t 1 2 = 0.693/ λ = 0.693/0.84 = 0.825 Therefore, the half life of the substance is 0.8 seconds. Example 2: Find the value of the decay constant of a radioactive substance having a half-life of 0.04 seconds. Solution:

  15. Half-Life & Doubling-Time

    f(t) = 300 ⋅2(t/12) or more generally f(t) = 300 ⋅2(t/T), where T is the doubling time. Exponential Growth A function that models exponential growth doubles in size after a characteristic time, T, called the doubling time. The exponential growth function can be written in the form f(t) = A ⋅ (2)t/T where

  16. Finding the Half-Life Practice

    1. A radioactive material went from 50g to 40g in 11 days. What approximate time in days will it take a sample of 50g to decay to 25g? 2. A certain bacteria population takes 195 mins for it to...

  17. 4.6: Exponential and Logarithmic Models

    Example. Estimate the value of point on the log scale above. The point appears to be half way between -2 and -1 in log value, so if is the value of this point, Rewriting in exponential form, Example. Place the number 6000 on a logarithmic scale. Solution. Since , this point would belong on the log scale about here:

  18. College Algebra

    Unit 1: Linear equations and inequalities 0/900 Mastery points Solving equations with one unknown Solutions to linear equations Multi-step linear inequalities Compound inequalities Modeling with linear equations and inequalities Absolute value equations Unit 2: Graphs and forms of linear equations 0/1100 Mastery points

  19. Lesson Radioactive decay problems

    Problem 1 The half-life of strontium-90 is 28 years. How much of a 10-g sample will remain after: a) 1 year? b) 10 years? Solution The general formula is M = , where is the initial mass; M is the current (remaining) mass, and "t" is time in years. So, the answers are (a) M = = 9.755 grams; (b) M = = 7.807 grams. Problem 2

  20. 6 Ways to Calculate Half Life

    1 What is half-life? The term "half-life" refers to the amount of time that half of the starting substance takes to decay or change. It's most often used in radioactive decay to figure out when a substance is no longer harmful to humans. [2] Elements like uranium and plutonium are most often studied with half-life in mind. 2

  21. half life

    Pre Algebra; Algebra; Equations. Basic (Linear) One-Step Addition; One-Step Subtraction; One-Step Multiplication; ... Word Problems. Age Problems; Distance Problems; Cost Problems; Investment Problems; Number Problems; ... half life. en. Related Symbolab blog posts. Middle School Math Solutions - Polynomials Calculator, Factoring Quadratics ...

  22. Half-Life Questions

    Correct Answer- (a.) 75 minutes Q3. How long does it take a 100.00g sample of As-81, with a half-life of 33 seconds, to decay to 6.25g? a.) 122 seconds b.) 101 seconds c.) 132 seconds d.) 22 seconds Correct Answer - (c.) 132 seconds Q4. What is the half-life of a radioactive isotope if a 500.0g sample decays to 62.5g in 24.3 hours? a.) 8.1 hours

  23. Half Life Chemistry Problems

    This chemistry video tutorial shows explains how to solve common half life radioactive decay problems. It shows you a simple technique to find the final amo...

  24. Half-Life Formula: Components and Applications

    Radioactive decay: Scientists use the half-life formula to describe the decay process of radioactive isotopes.It helps determine the rate of decay and predict how much of a substance will remain after a certain period. Carbon dating: Carbon dating relies on the half-life of carbon-14 (14 C) to estimate the age of organic materials. By measuring the ratio of carbon-14 to carbon-12 in a sample ...

  25. How China Broke One Man's Dreams

    And then, as the economy grew, life became much better, the opportunities were abundant, and there were, like, hundreds of millions of Chinese middle-class. And by 2012, very few Chinese were leaving.

  26. An Explosive Hearing in Trump's Georgia Election Case

    Fani T. Willis, the district attorney, defended her personal conduct as defense lawyers sought to disqualify her from the prosecution.