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Exponential Functions Word Problems Worksheet with Answers

Exponential functions are one of the most applicable mathematics topics to the real-world. Whether it’s calculating compound interest or exploring the exponential decay of a radioactive substance, there are just so many real-world applications that make excellent exponential functions word problems! But since word problems can be tricky, I have put together an exponential functions word problems worksheet with answers to help! 

Let’s get to the bottom of how to solve exponential functions word problems by exploring some real-world applications!

What is an Exponential Function Word Problem?

Exponential growth and decay happen  everywhere  in the real-world. From the bacteria growing on your toilet seat to the compound interest you earn in you bank account, there are so many applications of exponential growth and decay functions. So it makes sense that you will see a wide variety of applications when it comes to exponential function word problems. 

Exponential functions are a specific  type of equation  that can be used to model certain relationships that exist in the real world. In particular, exponential functions can be used to model exponential relationships, or relationships that involve rapid growth or decay by some factor. This is much different than linear functions, for example, which have a constant rate of change. 

Just like any other type of function, exponential functions are awesome because they allow us to make predictions about future amounts. And who doesn’t want to predict the future!? 

An exponential function word problem will ask you to do exactly this! Your goal will be to use the information provided to make a prediction about how much of something there will be in the future. 

For example, you could be tasked with finding anything ranging from the amount of a substance remaining, to the value of an investment, or the size of a population of bacteria after a certain amount of time.

pixel art bacteria

There are also a wide variety of exponential function word problems types. But in general, you will be given some initial amount, a growth rate or a decay rate, and an amount of time that has passed. In some cases you will also be given an exponential expression that you can use. If not, you will have to create your own exponential growth function or decay function!

How Do You Solve Exponential Function Word Problems?

The best way to solve exponential function word problems is by using a similar problem-solving process to what you would use in any other area of mathematics.

As a first step, I always recommend reading the question carefully. This sound obvious, but reading and re-reading the problem will help you to identify any key information that is given. This will help you determine exactly what values you will be using in your solution. 

For example, you should ask yourself questions such as:

  • Is the problem providing a growth rate or a decay rate? 
  • Do I know the initial value? Or is this what I am being tasked with determining? 
  • Is an exponential growth function or an exponential decay function being given? Do I have to write my own function?
  • What does the problem not tell me?

Once you have identified all of the key information that you are given, it is time to identify the unknown quantity that you are being asked to determine. As a tip, this is usually mentioned in the final sentence of the problem. It is important to take note of the units provided so that you can include those in your final answer as well.

Finally, your last step should be to reflect to select the appropriate problem-solving strategy. In some problems, exponential equations are provided that you can use. In other cases, you have to write your own. 

Most exponential function word problems will require you to make a substitution of some sort into an exponential function. After you perform this step, you will be required to perform a few basic operations in order to determine your answer.

I have covered exponential function word problems in the past on the Math By The Pixel YouTube channel , so I have linked a video walkthrough of a series of examples . Take a look at the following video as a primer before you download and complete the exponential functions word problems worksheet with answers linked below!

Download Your Exponential Functions Word Problems Worksheet with Answers

Below you will find an exponential functions word problems worksheet with answers. I constructed this worksheet by using the best problems that I have used throughout my teaching career to help students develop an understanding of exponential functions in the real-world.

I have hand picked a good mix of exponential growth and exponential decay problems across a wide variety of applications. For example, you will see exponential growth function examples such as compound interest and population growth. You will also see exponential decay examples such as depreciation of an asset and radioactive decay of a substance.

I also made sure to include problems that require a wide range of problem solving strategies. For example, you will be asked to use a function that is provided, as well as write and use your own! 

There are also a few problems that require you to identify the meaning of each parameter in a given exponential function. This will help deepen your understanding of how these important functions work!

Remember to check the answer keys provided to ensure that you fully understand each of these problems.

Download the worksheet by clicking below!

Practicing Exponential Functions Word Problems

Remember that there are a wide variety of exponential functions word problems that you can encounter. Practice is always key when it comes to mastering any mathematics concept. And exponential functions word problems are no different! 

Regular practice working with exponential expression word problems will help you become more comfortable with concepts like compound interest and radioactive decay.

When you complete these types of problems often you will start to notice patterns. And before long, you will start to develop a sense of mathematical intuition about how to solve specific real-world problems!

Did you find this exponential functions word problems worksheet with answers helpful? Share this post with a friend and subscribe to Math By The Pixel on YouTube for more helpful math content!

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How to Write an Exponential Function: Word Problems

Any function in the form \(y=ab^x\) is called an exponential function. Unlike linear functions where the growth rate is constant, exponential functions are characterized by the fact that the growth rate of the function is proportional to the function's current value. In this article, the method of writing an exponential function is explained step by step.

How to Write an Exponential Function: Word Problems

A step-by-step guide to Writing exponential function: word problems

Any function in the form \(y=ab^x\) is called an exponential function where \(a\) and \(b\) are fixed and \(x\) is an independent variable:

Variable \(y\): represents the output value

\(a\) represents the initial value

\(b\) represents the common ratio

\(x\) represents the number of times the original value has been multiplied by the common ratio

Unlike linear functions where the growth rate is constant, exponential functions are characterized by the fact that the growth rate of the function is proportional to the function’s current value and are used to model growth and decay processes, such as population growth, radioactive decay, and compound interest.

Follow the step-by-step procedure below to write an exponential function.

Step \(1\): Analyzing the problem and identifying the variables. At this stage, you should specify what the question is about and what the variables of the problem are.

Step \(2\): Using the variables, write the exponential function in the form of \(y=ab^x\).

Step \(3\): Determine the initial value (\(a\)) according to the information of the problem. \(a\) shows the value of the variable at the beginning of the time period.

Step \(4\): In this step, you must specify the common ratio (\(b\)). \(b\) is the coefficient by which the variable is multiplied in each time period.

Step \(5\): Specify the \(x\) variable that represents the number of time periods that have passed. For example, if in the problem it is said that a certain amount will be multiplied by \(4\) every month and after \(5\) months you will be asked for the amount, \(x=5\).

Step \(6\): Place the values of \(a\), \(b\) and \(x\) in the exponential function and simplify it.

Step \(7\): Check your work by inserting \(x\) values and seeing if they match the problem statement

Step \(8\): Write the final answer in one sentence

Writing exponential function: word problems-Example 1:

If a bank’s profits start at \($60,000\) and increase by \(25%\) each year, how much will the profits be after \(4\) years?

This problem is solved by using the exponential function. In the said problem, the profit of the company starts from \(60,000\) dollars and increases by \(25%\) every year, the profit is multiplied by \(1.2\) every year. With this information, we write the exponential function:

\(y = 60000 * 1.25^x\)

In this function, \(x\) is the number of past years and \(y\) represents the profit after \(x\) years.

The profit amount after \(4\) years is obtained by inserting \(x = 4\) in the function: \(y = 40000 * 1.25^4 = 60000 * 1.25 * 1.25 * 1.25 * 1.25 = 146484\)

by: Effortless Math Team about 1 year ago (category: Articles )

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Mastering Grade 6 Math Word Problems The Ultimate Guide to Tackling 6th Grade Math Word Problems

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WORD PROBLEMS ON EXPONENTIAL FUNCTIONS

Every exponential function will be in the form of y = ab x

Here a = initial value and b = base.

Based on the value of b, we can see two types of exponential function.

  • Exponential growth, if b > 1
  • Exponential decay, if 0 < b < 1

Some times exponential functions will be in the form

Here r is the growth or decay factor.

If the function is changes exponentially, then

Problem 1 :

A cup of green tea contains 35 milligrams of caffeine. The average teen can eliminate approximately 12.5% of the caffeine from their system per hour.

a) Write an exponential function to represent the amount of caffeine remaining after drinking a cup of green tea.

b) Estimate the amount of caffeine in a teenager’s body 3 hours after drinking a cup of green tea.

a)  a = 35

Per hour 12.5 % of caffeine is being eliminated each hour.

then, r = 12.5

y = 35(1 - 12.5%) x  

Here x is the number of hours.

y = 35(1 - 0.125) x  

y = 35(0.875) x  

b) Number of hours = 3

y = 35(0.875) 3  

= 35(0.6699) 

So, the amount caffeine after 3 hours is 23.45 milligrams

Problem 2 :

From 1990 to 2000, the population of California can be modeled by

P = 29,816,591(1.0128) t

where t is the number of years since 1990. Estimate the population in 2007.

To estimate the population in 2007, we have to apply t as 17.

P = 29,816,591(1.0128) 17

= 37013551.9

Approximately  37013552.

So, the required population at 2007 is 37013552..

Problem 3 :

You buy a new car for $22,500. The value of the car decreases by 25% each year.

Write an exponential decay model giving the car's value V (in dollars) after t years. What is the value of the car after three years?

Initial value = 22500

Decreasing rate = 25%

y = 22500(1 - 25%) x  

= 22500(1- 0.25) x  

= 22500(0.75) x  

Problem 4 :

A virus spreads through a network of computers such that each minute, 25% more computers are infected. If the virus began at only one computer, find the model for this situation and find the number of computers affected after 40 minutes.

Number of virus in the initial stage = 1

Increasing factor = 25%

y = 1(1 + 25%) x  

= 1(1 + 0.25) x  

= 1(1.25) x  

The total number of computer after 40 minutes.

= 1(1.25) 40

Problem 5 :

A radioactive substance decays exponentially. A scientist begins with 100 milligrams of a radioactive substance. After 35 hours, 50 mg of the substance remains. How many milligrams will remain after 54 hours?

Exponential function for growth or decay

A(t) = ae rt ---(1)

Here A(t) is the quantity of substance after t hours.

Initial value = 100

100 = a(e) r(0)  

100 = a(e) 0

Applying the value of a in (1), we get

A(t) = 100e rt

Here t = 35, A(t) = 50

50 = 100e r(35)

1/2 =  e r(35)

ln (1/2) = 35r

r = ln(0.5)/35

r = -0.693/35

Approximately r = -0.02

Rate of decreasing = 2%

Quantity of substance after 54 hours.

A(t) = 100e (-0.02) 54

= 100e (-1.08)

Approximately 34 milligrams.

Problem 6 :

A radioactive substance decays exponentially. A scientist begins with 110 milligrams of a radioactive substance. After 31 hours, 55 mg of the substance remains. How many milligrams will remain after 42 hours?

Initial value = 110

110 = a(e) r(0)  

110 = a(e) 0

A(t) = 110e rt

Here t = 31, A(t) = 55

55 = 110e r(31)

1/2 =  e r(31)

ln (1/2) = 31r

r = ln(0.5)/31

r = -0.693/31

Approximately r = -0.022

Rate of decreasing = 2.2%

Quantity of substance after 42 hours.

A(t) = 110e (-0.022) 42

= 110e (-0.924)

= 110/2.519

Approximately 44 milligrams.

Problem 7 :

A house was valued at $110,000 in the year 1985. The value appreciated to $145,000 by the year 2005. What was the annual growth rate between 1985 and 2005?

Assume that the house value continues to grow by the same percentage. What did the value equal in the year 2010?

By observing the given situation, the required function will be exponential growth function.

P(t) = a(1 + r%) x  

At 1985, the population is  $110,000

110000 = a(1 + r%) 0  

Applying the initial value in (1), we get

P(t) = 110000(1 + r%) x  

In the year 2005, the value of x = 20

145000 = 110000(1 + r%) 20  

(1 + r%) 20   = 145000/110000

(1 + r%) 20  = 1.318

1 + r% = (1.318) 1/20

1 + r% = (1.318) 0.05

r% = 1.014 - 1

In the year 2010, the value of x = 25

P(t) = 110000(1 + 1.4%) 25  

= 110000(1.416)

So, the population at 2010 is  155760.

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EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS

In this section, we are going to see how to solve word problems on exponential growth and decay.

Before look at the problems, if you like to learn about exponential growth and decay,

please click here

Problem 1 :

David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007 ?

Number of years between 1999 and 2007 is 

n  =  2007 - 1999

n  =  8

No. of stores in the year 2007  =   P(1+r)ⁿ

Substitute P = 200, r = 8% or 0.08 and n = 8.

No. of stores in the year 2007  =  200(1 + 0.08) 8

No. of stores in the year 2007  =  200(1.08 ) 8

No. of stores in the year 2007  =  200(1.8509)

No. of stores in the year 2007  =  370.18

So, the number of stores in the year 2007 is about 370.

Problem 2 :

You invest $2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?

We have to use the formula given below to know the value of the investment after 3 years. 

A  =  Pe rt

Substitute 

P  =  2500

r  =  10% or 0.1

t  =  10

e  =  2.71828

Then, we have 

A = 2500(2.71828) (0.1)10

A = 6795.70

So, the value of the investment after 10 years is $6795.70.

Problem 3 :

Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours ?

Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100. 

We have to use the formula given below to find the percent of substance after 6 hours. 

A  =  P(1 + r) n

P  =  100

r  =  -3.5% or -0.035

t  =  6

(Here, the value of "r" is taken in negative sign. because the substance decays)

A  =  100(1-0.035) 6

A  =  100(0.935 ) 6

A  =  100(0.8075)

A  =  80.75

Because  the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75% 

Problem 4 :

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria  present in the culture initially, how many bacteria will be present at the end of 8th hour?

Note that the number of bacteria present in the culture doubles at the end of  successive hours.

Since it grows at the constant ratio "2", the growth is based is on geometric progression. 

We have to use the formula given below to find the no. of bacteria present at the end of 8th hour. 

A  =  ab x

a  =  30

b  =  2

x  =  8

Then, we have

A  =  30(2 8 )

A  =  30(256 )

A  =  7680

So, the number of bacteria at the end of 8th hour is 7680.

Problem 5 :

A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself ?

Let "P" be the amount invested initially.  From the given information, P becomes 2P in 3 years. 

Since the investment is in compound interest, for the 4th year, the principal will be 2P.  And 2P becomes 4P (it doubles itself) in the next 3 years. 

Therefore, at the end of 6 years accumulated value will be 4P.  So, the amount deposited will amount to 4 times itself in 6 years.

Related Topics

Doubling-Time Growth Formula

Half-Life Decay Formula

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solving exponential function word problems

How Do You Solve a Word Problem Using an Exponential Function?

Word problems let you see math in the real world! This tutorial shows you how to create a table and identify a pattern from the word problem. Then you can see how to create an exponential function from the data and solve the function to get your answer!

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Exponential-Growth Word Problems

Log Probs Expo Growth Expo Decay

What is the formula for exponential growth?

Exponential growth word problems work off the exponential-growth formula, A  =  Pe rt , where A is the ending amount of whatever you're dealing with (for example, money sitting in an investment, or bacteria growing in a petri dish), P is the beginning amount of that same whatever, r is the growth constant, and t is time.

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Exponential Growth and Decay on MathHelp.com

Exponential Growth and Decay

The exponential-growth formula A  =  Pe rt is related to the compound-interest formula , and represents the case of the interest being compounded "continuously".

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 growth constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

How do you solve exponential growth problems?

To solve exponential growth word problems, you may be plugging one value into the exponential-growth equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the doubling-time information to find the growth 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 an exponential growth problem?

  • A biologist is researching a newly-discovered species of bacteria. At time t  = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant k for the bacteria? (Round k to two decimal places.)

For this exercise, the units on time t will be hours, because the growth is being measured in terms of hours.

The beginning amount P is the amount at time t  = 0 , so, for this problem, P  = 100 .

The ending amount is A  = 450 at t  = 6 hours.

The only variable I don't have a value for is the growth constant k , which also happens to be what I'm looking for. So I'll plug all the known values into the exponential-growth formula, and then solve for the growth constant:

450 = 100 e 6 k

4.5 = e 6 k

ln(4.5) = 6 k

ln(4.5) / 6 = k = 0.250679566129...

They want me to round my decimal value (found by punching keys on my calculator) to two decimal places. So my answer is:

Many math classes, math books, and math instructors leave off the units for the growth and decay rates. However, if you see this topic again in chemistry or physics, you will probably be expected to use proper units ("growth-decay constant ÷ time"). If I had included this information in my solution above, my answer would have been " k  = 0.25 /hour". I doubt that your math class will even mention this, let alone require that you include this. Still, it's not a bad idea to get into the habit now of checking and reporting your units.

Note that the constant was positive, because it was a growth constant. If I had come up with a negative value for the growth constant, then I would have known to check my work to find my error(s).

  • A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half?

In this problem, I know that time t will be in hours, because they gave me growth in terms of hours. As a result, I'll convert "a day and a half" to "thirty-six hours", so my units match.

I know that the starting population is P  = 100 , and I need to find A at time t  = 36 .

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But what is the growth constant k ? And why do they tell me what the doubling time is?

They gave me the doubling time because I can use this to find the growth constant k . Then, once I have this constant, I can go on to answer the actual question.

So this exercise actually has two unknowns, the growth constant k and the ending amount A . I can use the doubling time to find the growth constant, at which point the only remaining value will be the ending amount, which is what they actually asked for. So first I'll find the constant.

If the initial population were, say, 100 , then, in 6.5 hours (being the specified doubling time), the population would be 200 . (It doesn't matter what starting value I pick for this part of my solution, as long as my ending value is twice as much. It's the "twice as much" that matters here, more than the number that it's twice of.) I'll set this up and solve for k :

200 = 100 e 6.5 k

2 = e 6.5 k

At this point, I need to use logs to solve:

ln(2) = 6.5 k

ln(2) / 6.5 = k

I could simplify this to a decimal approximation, but I won't, because I don't want to introduce round-off error if I can avoid it. So, for now, the growth constant will remain this "exact" value. (I might want to check this value quickly in my calculator, to make sure that this growth constant is positive, as it should be. If I have a negative value at this stage, I need to go back and check my work.)

Now that I have the growth constant, I can answer the actual question, which was "How many bacteria will there be in thirty-six hours?" This means using 100 for P , 36 for t , and the above expression for k . I plug these values into the formula, and then I simplify to find A :

A = 100 e 36(ln(2)/6.5) = 4647.75313957...

I will take the luxury of assuming that they don't want fractional bacteria (being the 0.7313957... part), so I'll round to the nearest whole number.

about 4648 bacteria

You can do a rough check of this answer, using the fact that exponential processes involve doubling times. The doubling time in this case is 6.5 hours, or between 6 and 7 hours.

If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, 800 in eighteen hours, 1600 in twenty-four hours, 3200 in thirty hours, and 6400 in thirty-six hours.

If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twenty-one hours, 1600 in twenty-eight hours, and 3200 in thirty-five hours.

The answer I got above, 4678 bacteria in thirty-six hours, fits nicely between these two estimates.

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Note: When you are given a nice, neat doubling time, another method for solving the exercise is to use a base of 2 . First, figure out how many doubling-times that you've been given. In the above case, this would start by noting that "a day and a half" is 36 hours, so we have:

36 ÷ 6.5 = 72/13

Use this as the power on 2 :

100 × 2 (72/13) = 4647.75314...

Not all algebra classes cover this method. If you're required to use the first method for every exercise of this type, then do so (in order to get the full points). Otherwise, this base- 2 trick can be a time-saver. And, yes, you'd use a base of 3 if you'd been given a tripling-time, a base of 4 for a quadrupling-time, etc.

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Mathematics LibreTexts

5.6: Application Problems with Exponential and Logarithmic Functions

  • Last updated
  • Save as PDF
  • Page ID 38600

  • Rupinder Sekhon and Roberta Bloom
  • De Anza College

Learning Objectives

In this section, you will:

  • review strategies for solving equations arising from exponential formulas
  • solve application problems involving exponential functions and logarithmic functions

STRATEGIES FOR SOLVING EQUATIONS THAT CONTAIN EXPONENTS

When solving application problems that involve exponential and logarithmic functions, we need to pay close attention to the position of the variable in the equation to determine the proper way solve the equation we investigate solving equations that contain exponents.

Suppose we have an equation in the form : value = coefficient(base) exponent

We consider four strategies for solving the equation:

STRATEGY A : If the coefficient, base, and exponent are all known, we only need to evaluate the expression for coefficient(base) exponent to evaluate its value.

STRATEGY B: If the variable is the coefficient, evaluate the expression for (base) exponent . Then it becomes a linear equation which we solve by dividing to isolate the variable.

STRATEGY C : If the variable is in the exponent, use logarithms to solve the equation.

STRATEGY D: If the variable is not in the exponent, but is in the base, use roots to solve the equation.

Below we examine each strategy with one or two examples of its use.

Example \(\PageIndex{1}\)

Suppose that a stock’s price is rising at the rate of 7% per year, and that it continues to increase at this rate. If the value of one share of this stock is $43 now, find the value of one share of this stock three years from now.

The problem tells us that \(a\) = 43 and \(r\) = 0.07, so \(b = 1+ r = 1+ 0.07 = 1.07\)

Therefore, function is \(y = 43(1.07)^t\).

In this case we know that \(t\) = 3 years, and we need to evaluate \(y\) when \(t\) = 3.

At the end of 3 years, the value of this one share of this stock will be

\[y=43(1.07)^{3}=\$ 52.68 \nonumber \]

Example \(\PageIndex{2}\)

The value of a new car depreciates (decreases) after it is purchased. Suppose that the value of the car depreciates according to an exponential decay model. Suppose that the value of the car is $12000 at the end of 5 years and that its value has been decreasing at the rate of 9% per year. Find the value of the car when it was new.

The function is \(y = a(0.91)^t\)

In this case we know that when \(t\) = 5, then \(y\) = 12000; substituting these values gives

\[12000 = a(0.91)^5 \nonumber \]

We need to solve for the initial value a, the purchase price of the car when new.

First evaluate (0.91) 5 ; then solve the resulting linear equation to find \(a\).

\[ 1200 = a(0.624) \nonumber \]

\(a=\frac{12000}{0.624} = \$ 19,230.77\); The car's value was $19,230.77 when it was new.

Example \(\PageIndex{3}\)

A national park has a population of 5000 deer in the year 2016. Conservationists are concerned because the deer population is decreasing at the rate of 7% per year. If the population continues to decrease at this rate, how long will it take until the population is only 3000 deer?

\(r\) = -0.07 and \(b = 1+r = 1+(-0.07) = 0.93\) and the initial population is \(a\) = 5000

The exponential decay function is \(y = 5000(0.93)^t\)

To find when the population will be 3000, substitute \(y\) = 3000

\[ 3000 = 5000(0.93)^t \nonumber \]

Next, divide both sides by 5000 to isolate the exponential expression

\[\begin{array}{l} \frac{3000}{5000}=\frac{5000}{5000}(0.93)^{2} \\ 0.6=0.93^{t} \end{array} \nonumber \]

Rewrite the equation in logarithmic form; then use the change of base formula to evaluate.

\[t=\log _{0.93}(0.6) \nonumber \]

\(t = \frac{\ln(0.6)}{\ln(0.93)}=7.039\) years; After 7.039 years, there are 3000 deer.

Note: In Example \(\PageIndex{3}\), we needed to state the answer to several decimal places of precision to remain accurate. Evaluating the original function using a rounded value of \(t\) = 7 years gives a value that is close to 3000, but not exactly 3000.

\[y=5000(0.93)^{7}=3008.5 \text { deer } \nonumber \]

However using \(t\) = 7.039 years produces a value of 3000 for the population of deer

\[ y=5000(0.93)^{7.039}=3000.0016 \approx 3000 \text { deer } \nonumber \]

Example \(\PageIndex{4}\)

A video posted on YouTube initially had 80 views as soon as it was posted. The total number of views to date has been increasing exponentially according to the exponential growth function \(y = 80e^{0.2t}\), where \(t\) represents time measured in days since the video was posted. How many days does it take until 2500 people have viewed this video?

Let \(y\) be the total number of views \(t\) days after the video is initially posted. We are given that the exponential growth function is \(y = 80e^{0.2t}\) and we want to find the value of \(t\) for which \(y\) = 2500. Substitute \(y\) = 2500 into the equation and use natural log to solve for \(t\).

\[2500 = 80e^{0.12t} \nonumber \]

Divide both sides by the coefficient, 80, to isolate the exponential expression.

\[\begin{array}{c} \frac{2500}{80}=\frac{80}{80} e^{0.12 t} \\ 31.25=e^{0.12 t} \end{array} \nonumber \]

Rewrite the equation in logarithmic form

\[ 0.12t = \ln(31.25) \nonumber \]

Divide both sides by 0.04 to isolate \(t\); then use your calculator and its natural log function to evaluate the expression and solve for \(t\).

\[\begin{array}{l} \mathrm{t}=\frac{\ln (31.25)}{0.12} \\ \mathrm{t}=\frac{3.442}{0.12} \\ \mathrm{t} \approx 28.7 \text { days } \end{array} \nonumber \]

This video will have 2500 total views approximately 28.7 days after it was posted.

STRATEGY D: If the variable is not in the exponent, but is in the base, we use roots to solve the equation. It is important to remember that we only use logarithms when the variable is in the exponent.

Example \(\PageIndex{5}\)

A statistician creates a website to analyze sports statistics. His business plan states that his goal is to accumulate 50,000 followers by the end of 2 years (24 months from now). He hopes that if he achieves this goal his site will be purchased by a sports news outlet. The initial user base of people signed up as a result of pre-launch advertising is 400 people. Find the monthly growth rate needed if the user base is to accumulate to 50,000 users at the end of 24 months.

Let \(y\) be the total user base \(t\) months after the site is launched.

The growth function for this site is \(y = 400(1+r)^t\);

We don’t know the growth rate \(r\). We do know that when \(t\) = 24 months, then \(y\) = 50000.

Substitute the values of \(y\) and \(t\); then we need to solve for \(r\).

\[5000 = 400(1+r)^{24} \nonumber \]

Divide both sides by 400 to isolate (1+r) 24 on one side of the equation

\[\begin{array}{l} \frac{50000}{400}=\frac{400}{400}(1+r)^{24} \\ 125=(1+r)^{24} \end{array} \nonumber \]

Because the variable in this equation is in the base, we use roots:

\[\begin{array}{l} \sqrt[24]{125}=1+r \\ 125^{1 / 24}=1+r \\ 1.2228 \approx 1+r \\ 0.2228 \approx r \end{array} \nonumber \]

The website’s user base needs to increase at the rate of 22.28% per month in order to accumulate 50,000 users by the end of 24 months.

Example \(\PageIndex{6}\)

A fact sheet on caffeine dependence from Johns Hopkins Medical Center states that the half life of caffeine in the body is between 4 and 6 hours. Assuming that the typical half life of caffeine in the body is 5 hours for the average person and that a typical cup of coffee has 120 mg of caffeine.

  • Write the decay function.
  • Find the hourly rate at which caffeine leaves the body.
  • How long does it take until only 20 mg of caffiene is still in the body? www.hopkinsmedicine.org/psyc...fact_sheet.pdf

a. Let \(y\) be the total amount of caffeine in the body \(t\) hours after drinking the coffee.

Exponential decay function \(y = ab^t\) models this situation.

The initial amount of caffeine is \(a\) = 120.

We don’t know \(b\) or \(r\), but we know that the half- life of caffeine in the body is 5 hours. This tells us that when \(t\) = 5, then there is half the initial amount of caffeine remaining in the body.

\[\begin{array}{l} y=120 b^{t} \\ \frac{1}{2}(120)=120 b^{5} \\ 60=120 b^{5} \end{array} \nonumber \]

Divide both sides by 120 to isolate the expression \(b^5\) that contains the variable.

\[\begin{array}{l} \frac{60}{120}=\frac{120}{120} \mathrm{b}^{5} \\ 0.5=\mathrm{b}^{5} \end{array} \nonumber \]

The variable is in the base and the exponent is a number. Use roots to solve for \(b\):

\[\begin{array}{l} \sqrt[5]{0.5}=\mathrm{b} \\ 0.5^{1 / 5}=\mathrm{b} \\ 0.87=\mathrm{b} \end{array} \nonumber \]

We can now write the decay function for the amount of caffeine (in mg.) remaining in the body \(t\) hours after drinking a cup of coffee with 120 mg of caffeine

\[y=f(t)=120(0.87)^{t} \nonumber \]

b. Use \(b = 1 + r\) to find the decay rate \(r\). Because \(b = 0.87 < 1\) and the amount of caffeine in the body is decreasing over time, the value of \(r\) will be negative.

\[\begin{array}{l} 0.87=1+r \\ r=-0.13 \end{array} \nonumber \]

The decay rate is 13%; the amount of caffeine in the body decreases by 13% per hour.

c. To find the time at which only 20 mg of caffeine remains in the body, substitute \(y\) = 20 and solve for the corresponding value of \(t\).

\[\begin{array}{l} y=120(.87)^{t} \\ 20=120(.87)^{t} \end{array} \nonumber \]

Divide both sides by 120 to isolate the exponential expression.

\[\begin{array}{l} \frac{20}{120}=\frac{120}{120}\left(0.87^{t}\right) \\ 0.1667=0.87^{t} \end{array} \nonumber \]

Rewrite the expression in logarithmic form and use the change of base formula

\[\begin{array}{l} t=\log _{0.87}(0.1667) \\ t=\frac{\ln (0.1667)}{\ln (0.87)} \approx 12.9 \text { hours } \end{array} \nonumber \]

After 12.9 hours, 20 mg of caffeine remains in the body.

EXPRESSING EXPONENTIAL FUNCTIONS IN THE FORMS y = ab t and y = ae kt

Now that we’ve developed our equation solving skills, we revisit the question of expressing exponential functions equivalently in the forms \(y = ab^t\) and \(y = ae^{kt}\)

We’ve already determined that if given the form \(y = ae^{kt}\), it is straightforward to find \(b\).

Example \(\PageIndex{7}\)

For the following examples, assume \(t\) is measured in years.

  • Express \(y = 3500e^{0.25t}\) in form \(y = ab^t\) and find the annual percentage growth rate.
  • Express \(y = 28000e^{-0.32t}\) in form \(y = ab^t\) and find the annual percentage decay rate.

a. Express \(y = 3500e^{0.25t}\) in the form \(y = ab^t\)

\[\begin{array}{l} y=a e^{k t}=a b^{t} \\ a\left(e^{k}\right)^{t}=a b^{t} \end{array} \nonumber \]

Thus \(e^k=b\)

In this example \(b=e^{0.25} \approx 1.284\)

We rewrite the growth function as y = 3500(1.284 t )

To find \(r\), recall that \(b = 1+r\) \[\begin{aligned} &1.284=1+r\\ &0.284=\mathrm{r} \end{aligned} \nonumber \]

The continuous growth rate is \(k\) = 0.25 and the annual percentage growth rate is 28.4% per year.

b. Express \(y = 28000e^{-0.32t}\) in the form \(y = ab^t\)

In this example \(\mathrm{b}=e^{-0.32} \approx 0.7261\)

We rewrite the growth function as y = 28000(0.7261 t )

To find \(r\), recall that \(b = 1+r\) \[\begin{array}{l} 0.7261=1+r \\ 0.2739=r \end{array} \nonumber \]

The continuous decay rate is \(k\) = -0.32 and the annual percentage decay rate is 27.39% per year.

In the sentence, we omit the negative sign when stating the annual percentage decay rate because we have used the word “decay” to indicate that r is negative.

Example \(\PageIndex{8}\)

  • Express \(y = 4200 (1.078)^t\) in the form \(y =ae^{kt}\)
  • Express \(y = 150 (0.73)^t\) in the form \(y =ae^{kt}\)

a. Express \(y = 4200 (1.078)^t\) in the form \(y =ae^{kt}\)

\[\begin{array}{l} \mathrm{y}=\mathrm{a} e^{\mathrm{k} t}=\mathrm{ab}^{\mathrm{t}} \\ \mathrm{a}\left(e^{\mathrm{k}}\right)^{\mathrm{t}}=\mathrm{ab}^{\mathrm{t}} \\ e^{\mathrm{k}}=\mathrm{b} \\ e^{k}=1.078 \end{array} \nonumber \]

Therefore \(\mathrm{k}=\ln 1.078 \approx 0.0751\)

We rewrite the growth function as \(y = 3500e^{0.0751t}\)

b. Express \(y =150 (0.73)^t\) in the form \(y = ae^{kt}\)

\[\begin{array}{l} y=a e^{k t}=a b^{t} \\ a\left(e^{k}\right)^{t}=a b^{t} \\ e^{k}=b \\ e^{k}=0.73 \end{array} \nonumber \]

Therefore \(\mathrm{k}=\ln 0.73 \approx-0.3147\)

We rewrite the growth function as \(y = 150e^{-0.3147t}\)

AN APPLICATION OF A LOGARITHMIC FUNCTON

Suppose we invest $10,000 today and want to know how long it will take to accumulate to a specified amount, such as $15,000. The time \(t\) needed to reach a future value \(y\) is a logarithmic function of the future value: \(t = g(y)\)

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    Solution : Number of years between 1999 and 2007 is n = 2007 - 1999 n = 8 No. of stores in the year 2007 = P (1+r)ⁿ Substitute P = 200, r = 8% or 0.08 and n = 8. No. of stores in the year 2007 = 200 (1 + 0.08)8 No. of stores in the year 2007 = 200 (1.08)8 No. of stores in the year 2007 = 200 (1.8509) No. of stores in the year 2007 = 370.18

  11. How Do You Solve a Word Problem Using an Exponential Function?

    How Do You Solve a Word Problem Using an Exponential Function? Note: Word problems let you see math in the real world! This tutorial shows you how to create a table and identify a pattern from the word problem. Then you can see how to create an exponential function from the data and solve the function to get your answer! Keywords: problem

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

    Exponential growth word problems work off the exponential-growth formula, A = Pert, where A is the ending amount of whatever you're dealing with (for example, money sitting in an investment, or bacteria growing in a petri dish), P is the beginning amount of that same whatever, r is the growth constant, and t is time. MathHelp.com

  13. Algebra

    Given the function f (x) = ( 1 5)x f ( x) = ( 1 5) x evaluate each of the following. Sketch each of the following. Sketch the graph of f (x) = e−x f ( x) = e − x. Solution. Sketch the graph of f (x) = ex−3 +6 f ( x) = e x − 3 + 6. Solution. Here is a set of practice problems to accompany the Exponential Functions section of the ...

  14. Mastering 3-1 Word Problem Practice with Exponential Functions: Answer

    Solving Word Problems Involving Exponential Functions. Exponential functions are mathematical functions that involve a base raised to a power. These functions have a unique characteristic of rapid growth or decay, making them useful in modeling various real-world phenomena. One common application of exponential functions is solving word problems.

  15. Solving Problems Involving Exponential Functions

    Example: Solve First, we have to cancel the coefficient behind the exponential function. Therefore, we divide both sides by 5: We need to isolate x, to do so, we have to cancel the power of x: In some other cases, one has to write an exponential function in a different base. Example: Convert to base We substitute into the original equation:

  16. 5.6: Application Problems with Exponential and Logarithmic Functions

    y = 10000(1.05)t y = 10000 ( 1.05) t. We divide both sides by 10000 to isolate the exponential expression on one side. y 10000 = 1.05t y 10000 = 1.05 t. Next we rewrite this in logarithmic form to express time as a function of the accumulated future value. We'll use function notation and call this function g(y) g ( y).

  17. Exponential expressions word problems (algebraic)

    If we are growing, well, I almost made a mistake there, it's 1.3. So here you go, 1.3. One plus 0.3 is 1.3. So once again, take a hard look at this right over here because it's going to be something that you see a lot with exponential functions. When you grow by 30%, that means you keep your 100% that you had before, and then you add another 30%.

  18. Word Problems Calculator

    An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.