Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Mathematics LibreTexts

4.7: Exponential and Logarithmic Equations

  • Last updated
  • Save as PDF
  • Page ID 64854

Learning Objectives

  • Use like bases to solve exponential equations.
  • Use logarithms to solve exponential equations.
  • Use the definition of a logarithm to solve logarithmic equations.
  • Use the one-to-one property of logarithms to solve logarithmic equations.
  • Solve applied problems involving exponential and logarithmic equations.

In 1859, an Australian landowner named Thomas Austin released \(24\) rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Seven rabbits in front of a brick building.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers \(b\), \(S\), and \(T\), where \(b>0\), \(b≠1\), \(b^S=b^T\) if and only if \(S=T\).

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation \(3^{4x−7}=\dfrac{3^{2x}}{3}\). To solve for \(x\), we use the division property of exponents to rewrite the right side so that both sides have the common base, \(3\). Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for \(x\):

\[\begin{align*} 3^{4x-7}&= \dfrac{3^{2x}}{3}\\ 3^{4x-7}&= \dfrac{3^{2x}}{3^1} \qquad &&\text{Rewrite 3 as } 3^1\\ 3^{4x-7}&= 3^{2x-1} \qquad &&\text{Use the division property of exponents}\\ 4x-7&= 2x-1 \qquad &&\text{Apply the one-to-one property of exponents}\\ 2x&= 6 \qquad &&\text{Subtract 2x and add 7 to both sides}\\ x&= 3 \qquad &&\text{Divide by 3} \end{align*}\]

USING THE ONE-TO-ONE PROPERTY OF EXPONENTIAL FUNCTIONS TO SOLVE EXPONENTIAL EQUATIONS

For any algebraic expressions \(S\) and \(T\), and any positive real number \(b≠1\),

\[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]

How to: Given an exponential equation with the form \(b^S=b^T\), where \(S\) and \(T\) are algebraic expressions with an unknown, solve for the unknown.

  • Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form \(b^S=b^T\).
  • Use the one-to-one property to set the exponents equal.
  • Solve the resulting equation, \(S=T\), for the unknown.

Example \(\PageIndex{1}\): Solving an Exponential Equation with a Common Base

Solve \(2^{x−1}=2^{2x−4}\).

\[\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{1}\)

Solve \(5^{2x}=5^{3x+2}\).

\(x=−2\)

Rewrite Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation \(256=4^{x−5}\). We can rewrite both sides of this equation as a power of \(2\). Then we apply the rules of exponents, along with the one-to-one property, to solve for \(x\):

\[\begin{align*} 256&= 4^{x-5}\\ 2^8&= {(2^2)}^{x-5} \qquad &&\text{Rewrite each side as a power with base 2}\\ 2^8&= 2^{2x-10} \qquad &&\text{Use the one-to-one property of exponents}\\ 8&= 2x-10 \qquad &&\text{Apply the one-to-one property of exponents}\\ 18&= 2x \qquad &&\text{Add 10 to both sides}\\ x&= 9 \qquad &&\text{Divide by 2} \end{align*}\]

How to: Given an exponential equation with unlike bases, use the one-to-one property to solve it.

  • Rewrite each side in the equation as a power with a common base.

Example \(\PageIndex{2}\): Solving Equations by Rewriting Them to Have a Common Base

Solve \(8^{x+2}={16}^{x+1}\).

\[\begin{align*} 8^{x+2}&= {16}^{x+1}\\ {(2^3)}^{x+2}&= {(2^4)}^{x+1} \qquad &&\text{Write 8 and 16 as powers of 2}\\ 2^{3x+6}&= 2^{4x+4} \qquad &&\text{To take a power of a power, multiply exponents}\\ 3x+6&= 4x+4 \qquad &&\text{Use the one-to-one property to set the exponents equal}\\ x&= 2 \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{2}\)

Solve \(5^{2x}={25}^{3x+2}\).

\(x=−1\)

Example \(\PageIndex{3}\): Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve \(2^{5x}=\sqrt{2}\).

\[\begin{align*} 2^{5x}&= 2^{\frac{1}{2}} \qquad &&\text{Write the square root of 2 as a power of 2}\\ 5x&= \dfrac{1}{2} \qquad &&\text{Use the one-to-one property}\\ x&= \dfrac{1}{10} \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{3}\)

Solve \(5^x=\sqrt{5}\).

\(x=\dfrac{1}{2}\)

Q&A: Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.

Example \(\PageIndex{4}\): Solving an Equation with Positive and Negative Powers

Solve \(3^{x+1}=−2\).

This equation has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive.

Figure \(\PageIndex{2}\) shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.

Exercise \(\PageIndex{4}\)

Solve \(2^x=−100\).

The equation has no solution.

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since \(\log(a)=\log(b)\) is equivalent to \(a=b\), we may apply logarithms with the same base on both sides of an exponential equation.

How to: Given an exponential equation in which a common base cannot be found, solve for the unknown

  • If one of the terms in the equation has base 10, use the common logarithm.
  • If none of the terms in the equation has base 10, use the natural logarithm.
  • Use the rules of logarithms to solve for the unknown.

Example \(\PageIndex{5}\): Solving an Equation Containing Powers of Different Bases

Solve \(5^{x+2}=4^x\).

\[\begin{align*} 5^{x+2}&= 4^x \qquad &&\text{There is no easy way to get the powers to have the same base}\\ \ln5^{x+2}&= \ln4^x \qquad &&\text{Take ln of both sides}\\ (x+2)\ln5&= x\ln4 \qquad &&\text{Use laws of logs}\\ x\ln5+2\ln5&= x\ln4 \qquad &&\text{Use the distributive law}\\ x\ln5-x\ln4&= -2\ln5 \qquad &&\text{Get terms containing x on one side, terms without x on the other}\\ x(\ln5-\ln4)&= -2\ln5 \qquad &&\text{On the left hand side, factor out an x}\\ x\ln \left (\dfrac{5}{4} \right )&= \ln \left (\dfrac{1}{25} \right ) \qquad &&\text{Use the laws of logs}\\ x&=\dfrac{\ln \left (\dfrac{1}{25} \right )}{\ln \left (\dfrac{5}{4} \right )} \qquad &&\text{Divide by the coefficient of x} \end{align*}\]

Exercise \(\PageIndex{5}\)

Solve \(2^x=3^{x+1}\).

\(x=\dfrac{\ln3}{\ln \left (\dfrac{2}{3} \right )}\)

Q&A: Is there any way to solve \(2^x=3^x\)?

Yes. The solution is \(0\).

Equations Containing \(e\)

One common type of exponential equations are those with base \(e\). This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base \(e\) on either side, we can use the natural logarithm to solve it.

How to: Given an equation of the form \(y=Ae^{kt}\), solve for \(t\).

  • Divide both sides of the equation by \(A\).
  • Apply the natural logarithm of both sides of the equation.
  • Divide both sides of the equation by \(k\).

Example \(\PageIndex{6}\): Solve an Equation of the Form \(y = Ae^{kt}\)

Solve \(100=20e^{2t}\).

\[\begin{align*} 100&= 20e^{2t}\\ 5&= e^{2t} \qquad &&\text{Divide by the coefficient of the power}\\ \ln5&= 2t \qquad &&\text{Take ln of both sides. Use the fact that } ln(x) \text{ and } e^x \text{ are inverse functions}\\ t&= \dfrac{\ln5}{2} \qquad &&\text{Divide by the coefficient of t} \end{align*}\]

Using laws of logs, we can also write this answer in the form \(t=\ln\sqrt{5}\). If we want a decimal approximation of the answer, we use a calculator.

Exercise \(\PageIndex{6}\)

Solve \(3e^{0.5t}=11\).

\(t=2\ln \left (\dfrac{11}{3} \right )\) or \(\ln{ \left (\dfrac{11}{3} \right )}^2\)

Q&A: Does every equation of the form \(y=Ae^{kt}\) have a solution?

No. There is a solution when \(k≠0\),and when \(y\) and \(A\) are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is \(2=−3e^t\).

Example \(\PageIndex{7}\): Solving an Equation That Can Be Simplified to the Form \(y=Ae^{kt}\)

Solve \(4e^{2x}+5=12\).

\[\begin{align*} 4e^{2x}+5&= 12\\ 4e^{2x}&= 7 \qquad &&\text{Combine like terms}\\ e^{2x}&= \dfrac{7}{4} \qquad &&\text{Divide by the coefficient of the power}\\ 2x&= \ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Take ln of both sides}\\ x&= \dfrac{1}{2}\ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{7}\)

Solve \(3+e^{2t}=7e^{2t}\).

\(t=\ln \left (\dfrac{1}{\sqrt{2}} \right )=−\dfrac{1}{2}\ln(2)\)

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Example \(\PageIndex{8}\): Solving Exponential Functions in Quadratic Form

Solve \(e^{2x}−e^x=56\).

\[\begin{align*} e^{2x}-e^x&= 56\\ e^{2x}-e^x-56&= 0 \qquad &&\text{Get one side of the equation equal to zero}\\ (e^x+7)(e^x-8)&= 0 \qquad &&\text{Factor by the FOIL method}\\ e^x+7&= 0 \qquad &&\text{or} \\ e^x-8&= 0 \qquad &&\text{If a product is zero, then one factor must be zero}\\ e^x&= -7 \qquad &&\text{or} \\ e^x&= 8 \qquad &&\text{Isolate the exponentials}\\ e^x&= 8 \qquad &&\text{Reject the equation in which the power equals a negative number}\\ x&= \ln8 \qquad &&\text{Solve the equation in which the power equals a positive number} \end{align*}\]

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation \(e^x=−7\) because a positive number never equals a negative number. The solution \(\ln(−7)\) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Exercise \(\PageIndex{8}\)

Solve \(e^{2x}=e^x+2\).

Q&A: Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation \({\log}_b(x)=y\) is equivalent to the exponential equation \(b^y=x\). We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation \({\log}_2(2)+{\log}_2(3x−5)=3\). To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x\):

\[\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\ {\log}_2(2(3x-5))&= 3 \qquad \text{Apply the product rule of logarithms}\\ {\log}_2(6x-10)&= 3 \qquad \text{Distribute}\\ 2^3&= 6x-10 \qquad \text{Apply the definition of a logarithm}\\ 8&= 6x-10 \qquad \text{Calculate } 2^3\\ 18&= 6x \qquad \text{Add 10 to both sides}\\ x&= 3 \qquad \text{Divide by 6} \end{align*}\]

USING THE DEFINITION OF A LOGARITHM TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expression \(S\) and real numbers \(b\) and \(c\),where \(b>0\), \(b≠1\),

\[\begin{align} {\log}_b(S)=c \text{ if and only if } b^c=S \end{align}\]

Example \(\PageIndex{9}\): Using Algebra to Solve a Logarithmic Equation

Solve \(2\ln x+3=7\).

\[\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad \text{Subtract 3}\\ \ln x&= 2 \qquad \text{Divide by 2}\\ x&= e^2 \qquad \text{Rewrite in exponential form} \end{align*}\]

Exercise \(\PageIndex{9}\)

Solve \(6+\ln x=10\).

Example \(\PageIndex{10}\): Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve \(2\ln(6x)=7\).

\[\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad \text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad \text{Use the definition of }\ln \\ x&= \dfrac{1}{6}e^{\left (\dfrac{7}{2} \right )} \qquad \text{Divide by 6} \end{align*}\]

Exercise \(\PageIndex{10}\)

Solve \(2\ln(x+1)=10\).

\(x=e^5−1\)

Example \(\PageIndex{11}\): Using a Graph to Understand the Solution to a Logarithmic Equation

Solve \(\ln x=3\).

\[\begin{align*} \ln x&= 3\\ x&= e^3 \qquad \text{Use the definition of the natural logarithm} \end{align*}\]

Figure \(\PageIndex{3}\) represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to \(20\). In other words \(e^3≈20\). A calculator gives a better approximation: \(e^3≈20.0855\).

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).

Exercise \(\PageIndex{11}\)

Use a graphing calculator to estimate the approximate solution to the logarithmic equation \(2^x=1000\) to \(2\) decimal places.

\(x≈9.97\)

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers \(x>0\), \(S>0\), \(T>0\) and any positive real number \(b\), where \(b≠1\),

\({\log}_bS={\log}_bT\) if and only if \(S=T\).

For example,

If \({\log}_2(x−1)={\log}_2(8)\), then \(x−1=8\).

So, if \(x−1=8\), then we can solve for \(x\),and we get \(x=9\). To check, we can substitute \(x=9\) into the original equation: \({\log}_2(9−1)={\log}_2(8)=3\). In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation \(\log(3x−2)−\log(2)=\log(x+4)\). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for \(x\):

\[\begin{align*} \log(3x-2)-\log(2)&= \log(x+4)\\ \log \left (\dfrac{3x-2}{2} \right )&= \log(x+4) \qquad \text{Apply the quotient rule of logarithms}\\ \dfrac{3x-2}{2}&= x+4 \qquad \text{Apply the one to one property of a logarithm}\\ 3x-2&= 2x+8 \qquad \text{Multiply both sides of the equation by 2}\\ x&= 10 \qquad \text{Subtract 2x and add 2} \end{align*}\]

To check the result, substitute \(x=10\) into \(\log(3x−2)−\log(2)=\log(x+4)\).

\[\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\ \log(28)-\log(2)&= \log(14)\\ \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad \text{The solution checks} \end{align*}\]

USING THE ONE-TO-ONE PROPERTY OF LOGARITHMS TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expressions \(S\) and \(T\) and any positive real number \(b\), where \(b≠1\),

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How to: Given an equation containing logarithms, solve it using the one-to-one property

  • Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form \({\log}_bS={\log}_bT\).
  • Use the one-to-one property to set the arguments equal.

Example \(\PageIndex{12}\): Solving an Equation Using the One-to-One Property of Logarithms

Solve \(\ln(x^2)=\ln(2x+3)\).

\[\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad \text{Use the one-to-one property of the logarithm}\\ x^2-2x-3&= 0 \qquad \text{Get zero on one side before factoring}\\ (x-3)(x+1)&= 0 \qquad \text{Factor using FOIL}\\ x-3&= 0 \qquad \text{or } x+1=0 \text{ If a product is zero, one of the factors must be zero}\\ x=3 \qquad \text{or} \\ x&= -11 \qquad \text{Solve for x} \end{align*}\]

There are two solutions: \(3\) or \(−1\). The solution \(−1\) is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Exercise \(\PageIndex{12}\)

Solve \(\ln(x^2)=\ln1\).

\(x=1\) or \(x=−1\)

Solving Applied Problems Using Exponential and Logarithmic Equations

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 . Table \(\PageIndex{1}\) 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:

\[\begin{align} A(t)&= A_0e^{\tfrac{\ln(0.5)}{T}t}\\ A(t)&= A_0e^{\tfrac{\ln(0.5)t}{T}}\\ A(t)&= A_0{(e^{\ln(0.5)})}^{\tfrac{t}{T}}\\ A(t)&= A_0{\left (\dfrac{1}{2}\right )}^{\tfrac{t}{T}}\\ \end{align}\]

  • \(A_0\) is the amount initially present
  • \(T\) is the half-life of the substance
  • \(t\) is the time period over which the substance is studied
  • \(y\) is the amount of the substance present after time \(t\)

Example \(\PageIndex{13}\): Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a \(1000\)-gram sample of uranium-235 to decay?

\[\begin{align*} y&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t}\\ 900&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{After } 10\% \text{ decays, 900 grams are left}\\ 0.9&= e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{Divide by 1000}\\ \ln(0.9)&= \ln \left (e^{\tfrac{\ln(0.5)}{703,800,000}t} \right ) \qquad \text{Take ln of both sides}\\ \ln(0.9)&= \dfrac{\ln(0.5)}{703,800,000}t \qquad \ln(e^M)=M\\ t&= 703,800,000\times \dfrac{\ln(0.9)}{\ln(0.5)} \qquad \text{years Solve for t}\\ t&\approx 106,979,777 \qquad \text{years} \end {align*} \]

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

Exercise \(\PageIndex{13}\)

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

\(t=703,800,000×\dfrac{\ln(0.8)}{\ln(0.5)}\)years ≈ 226,572,993 years.

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

  • Solving Logarithmic Equations
  • Solving Exponential Equations with Logarithms

Key Equations

Key concepts.

  • We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
  • When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example \(\PageIndex{1}\).
  • When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), and Example \(\PageIndex{4}\).
  • When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example \(\PageIndex{5}\).
  • We can solve exponential equations with base \(e\),by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example \(\PageIndex{6}\) and Example \(\PageIndex{7}\).
  • After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example \(\PageIndex{8}\).
  • When given an equation of the form \({\log}_b(S)=c\), where \(S\) is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation \(b^c=S\), and solve for the unknown. See Example \(\PageIndex{9}\) and Example \(\PageIndex{10}\).
  • We can also use graphing to solve equations with the form \({\log}_b(S)=c\). We graph both equations \(y={\log}_b(S)\) and \(y=c\) on the same coordinate plane and identify the solution as the x- value of the intersecting point. See Example \(\PageIndex{11}\).
  • When given an equation of the form \({\log}_bS={\log}_bT\), where \(S\) and \(T\) are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation \(S=T\) for the unknown. See Example \(\PageIndex{12}\).
  • Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example \(\PageIndex{13}\).
  • 6.6 Exponential and Logarithmic Equations
  • 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
  • 6.7 Exponential and Logarithmic Models
  • 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:

  • Use like bases to solve exponential equations.
  • Use logarithms to solve exponential equations.
  • Use the definition of a logarithm to solve logarithmic equations.
  • Use the one-to-one property of logarithms to solve logarithmic equations.
  • Solve applied problems involving exponential and logarithmic equations.

In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b , b , S , S , and T , T , where b > 0 , b ≠ 1 , b > 0 , b ≠ 1 , b S = b T b S = b T if and only if S = T . S = T .

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation 3 4 x − 7 = 3 2 x 3 . 3 4 x − 7 = 3 2 x 3 . To solve for x , x , we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x x :

Using the One-to-One Property of Exponential Functions to Solve Exponential Equations

For any algebraic expressions S and  T , S and  T , and any positive real number b ≠ 1 , b ≠ 1 ,

Given an exponential equation with the form b S = b T , b S = b T , where S S and T T are algebraic expressions with an unknown, solve for the unknown.

  • Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form b S = b T . b S = b T .
  • Use the one-to-one property to set the exponents equal.
  • Solve the resulting equation, S = T , S = T , for the unknown.

Solving an Exponential Equation with a Common Base

Solve 2 x − 1 = 2 2 x − 4 . 2 x − 1 = 2 2 x − 4 .

Solve 5 2 x = 5 3 x + 2 . 5 2 x = 5 3 x + 2 .

Rewriting Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation 256 = 4 x − 5 . 256 = 4 x − 5 . We can rewrite both sides of this equation as a power of 2. 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for x : x :

Given an exponential equation with unlike bases, use the one-to-one property to solve it.

  • Rewrite each side in the equation as a power with a common base.

Solving Equations by Rewriting Them to Have a Common Base

Solve 8 x + 2 = 16 x + 1 . 8 x + 2 = 16 x + 1 .

Solve 5 2 x = 25 3 x + 2 . 5 2 x = 25 3 x + 2 .

Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve 2 5 x = 2 . 2 5 x = 2 .

Solve 5 x = 5 . 5 x = 5 .

Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.

Solving an Equation with Positive and Negative Powers

Solve 3 x + 1 = −2. 3 x + 1 = −2.

This equation has no solution. There is no real value of x x that will make the equation a true statement because any power of a positive number is positive.

Figure 2 shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Solve 2 x = −100. 2 x = −100.

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since log ( a ) = log ( b ) log ( a ) = log ( b ) is equivalent to a = b , a = b , we may apply logarithms with the same base on both sides of an exponential equation.

Given an exponential equation in which a common base cannot be found, solve for the unknown.

  • If one of the terms in the equation has base 10, use the common logarithm.
  • If none of the terms in the equation has base 10, use the natural logarithm.
  • Use the rules of logarithms to solve for the unknown.

Solving an Equation Containing Powers of Different Bases

Solve 5 x + 2 = 4 x . 5 x + 2 = 4 x .

Solve 2 x = 3 x + 1 . 2 x = 3 x + 1 .

Is there any way to solve 2 x = 3 x ? 2 x = 3 x ?

Yes. The solution is 0. 0.

Equations Containing e

One common type of exponential equations are those with base e . e . This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e e on either side, we can use the natural logarithm to solve it.

Given an equation of the form y = A e k t , y = A e k t , solve for t . t .

  • Divide both sides of the equation by A . A .
  • Apply the natural logarithm of both sides of the equation.
  • Divide both sides of the equation by k . k .

Solve an Equation of the Form y = Ae kt

Solve 100 = 20 e 2 t . 100 = 20 e 2 t .

Using laws of logs, we can also write this answer in the form t = ln 5 . t = ln 5 . If we want a decimal approximation of the answer, we use a calculator.

Solve 3 e 0.5 t = 11. 3 e 0.5 t = 11.

Does every equation of the form y = A e k t y = A e k t have a solution?

No. There is a solution when k ≠ 0 , k ≠ 0 , and when y y and A A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is 2 = −3 e t . 2 = −3 e t .

Solving an Equation That Can Be Simplified to the Form y = Ae kt

Solve 4 e 2 x + 5 = 12. 4 e 2 x + 5 = 12.

Solve 3 + e 2 t = 7 e 2 t . 3 + e 2 t = 7 e 2 t .

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Solving Exponential Functions in Quadratic Form

Solve e 2 x − e x = 56. e 2 x − e x = 56.

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation e x = −7 e x = −7 because a positive number never equals a negative number. The solution ln ( −7 ) ln ( −7 ) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Solve e 2 x = e x + 2. e 2 x = e x + 2.

Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation log b ( x ) = y log b ( x ) = y is equivalent to the exponential equation b y = x . b y = x . We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation log 2 ( 2 ) + log 2 ( 3 x − 5 ) = 3. log 2 ( 2 ) + log 2 ( 3 x − 5 ) = 3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x : x :

For any algebraic expression S S and real numbers b b and c , c , where b > 0 , b ≠ 1 , b > 0 , b ≠ 1 ,

Using Algebra to Solve a Logarithmic Equation

Solve 2 ln x + 3 = 7. 2 ln x + 3 = 7.

Solve 6 + ln x = 10. 6 + ln x = 10.

Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve 2 ln ( 6 x ) = 7. 2 ln ( 6 x ) = 7.

Solve 2 ln ( x + 1 ) = 10. 2 ln ( x + 1 ) = 10.

Using a Graph to Understand the Solution to a Logarithmic Equation

Solve ln x = 3. ln x = 3.

Figure 3 represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to 20. In other words e 3 ≈ 20. e 3 ≈ 20. A calculator gives a better approximation: e 3 ≈ 20.0855. e 3 ≈ 20.0855.

Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2 x = 1000 2 x = 1000 to 2 decimal places.

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0 , x > 0 , S > 0 , S > 0 , T > 0 T > 0 and any positive real number b , b , where b ≠ 1 , b ≠ 1 ,

For example,

So, if x − 1 = 8 , x − 1 = 8 , then we can solve for x , x , and we get x = 9. x = 9. To check, we can substitute x = 9 x = 9 into the original equation: log 2 ( 9 − 1 ) = log 2 ( 8 ) = 3. log 2 ( 9 − 1 ) = log 2 ( 8 ) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation log ( 3 x − 2 ) − log ( 2 ) = log ( x + 4 ) . log ( 3 x − 2 ) − log ( 2 ) = log ( x + 4 ) . To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x : x :

To check the result, substitute x = 10 x = 10 into log ( 3 x − 2 ) − log ( 2 ) = log ( x + 4 ) . log ( 3 x − 2 ) − log ( 2 ) = log ( x + 4 ) .

For any algebraic expressions S S and T T and any positive real number b , b , where b ≠ 1 , b ≠ 1 ,

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

Given an equation containing logarithms, solve it using the one-to-one property.

  • Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form log b S = log b T . log b S = log b T .
  • Use the one-to-one property to set the arguments equal.

Solving an Equation Using the One-to-One Property of Logarithms

Solve ln ( x 2 ) = ln ( 2 x + 3 ) . ln ( x 2 ) = ln ( 2 x + 3 ) .

There are two solutions: 3 3 or −1. −1. The solution −1 −1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Solve ln ( x 2 ) = ln 1. ln ( x 2 ) = ln 1.

Solving Applied Problems Using Exponential and Logarithmic Equations

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 . Table 1 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:

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

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

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

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?

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

  • Solving Logarithmic Equations
  • Solving Exponential Equations with Logarithms

How can an exponential equation be solved?

When does an extraneous solution occur? How can an extraneous solution be recognized?

When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

For the following exercises, use like bases to solve the exponential equation.

4 − 3 v − 2 = 4 − v 4 − 3 v − 2 = 4 − v

64 ⋅ 4 3 x = 16 64 ⋅ 4 3 x = 16

3 2 x + 1 ⋅ 3 x = 243 3 2 x + 1 ⋅ 3 x = 243

2 − 3 n ⋅ 1 4 = 2 n + 2 2 − 3 n ⋅ 1 4 = 2 n + 2

625 ⋅ 5 3 x + 3 = 125 625 ⋅ 5 3 x + 3 = 125

36 3 b 36 2 b = 216 2 − b 36 3 b 36 2 b = 216 2 − b

( 1 64 ) 3 n ⋅ 8 = 2 6 ( 1 64 ) 3 n ⋅ 8 = 2 6

For the following exercises, use logarithms to solve.

9 x − 10 = 1 9 x − 10 = 1

2 e 6 x = 13 2 e 6 x = 13

e r + 10 − 10 = −42 e r + 10 − 10 = −42

2 ⋅ 10 9 a = 29 2 ⋅ 10 9 a = 29

− 8 ⋅ 10 p + 7 − 7 = −24 − 8 ⋅ 10 p + 7 − 7 = −24

7 e 3 n − 5 + 5 = −89 7 e 3 n − 5 + 5 = −89

e − 3 k + 6 = 44 e − 3 k + 6 = 44

− 5 e 9 x − 8 − 8 = −62 − 5 e 9 x − 8 − 8 = −62

− 6 e 9 x + 8 + 2 = −74 − 6 e 9 x + 8 + 2 = −74

2 x + 1 = 5 2 x − 1 2 x + 1 = 5 2 x − 1

e 2 x − e x − 132 = 0 e 2 x − e x − 132 = 0

7 e 8 x + 8 − 5 = −95 7 e 8 x + 8 − 5 = −95

10 e 8 x + 3 + 2 = 8 10 e 8 x + 3 + 2 = 8

4 e 3 x + 3 − 7 = 53 4 e 3 x + 3 − 7 = 53

8 e − 5 x − 2 − 4 = −90 8 e − 5 x − 2 − 4 = −90

3 2 x + 1 = 7 x − 2 3 2 x + 1 = 7 x − 2

e 2 x − e x − 6 = 0 e 2 x − e x − 6 = 0

3 e 3 − 3 x + 6 = −31 3 e 3 − 3 x + 6 = −31

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

log ( 1 100 ) = −2 log ( 1 100 ) = −2

log 324 ( 18 ) = 1 2 log 324 ( 18 ) = 1 2

For the following exercises, use the definition of a logarithm to solve the equation.

5 log 7 n = 10 5 log 7 n = 10

− 8 log 9 x = 16 − 8 log 9 x = 16

4 + log 2 ( 9 k ) = 2 4 + log 2 ( 9 k ) = 2

2 log ( 8 n + 4 ) + 6 = 10 2 log ( 8 n + 4 ) + 6 = 10

10 − 4 ln ( 9 − 8 x ) = 6 10 − 4 ln ( 9 − 8 x ) = 6

For the following exercises, use the one-to-one property of logarithms to solve.

ln ( 10 − 3 x ) = ln ( − 4 x ) ln ( 10 − 3 x ) = ln ( − 4 x )

log 13 ( 5 n − 2 ) = log 13 ( 8 − 5 n ) log 13 ( 5 n − 2 ) = log 13 ( 8 − 5 n )

log ( x + 3 ) − log ( x ) = log ( 74 ) log ( x + 3 ) − log ( x ) = log ( 74 )

ln ( − 3 x ) = ln ( x 2 − 6 x ) ln ( − 3 x ) = ln ( x 2 − 6 x )

log 4 ( 6 − m ) = log 4 3 m log 4 ( 6 − m ) = log 4 3 m

ln ( x − 2 ) − ln ( x ) = ln ( 54 ) ln ( x − 2 ) − ln ( x ) = ln ( 54 )

log 9 ( 2 n 2 − 14 n ) = log 9 ( − 45 + n 2 ) log 9 ( 2 n 2 − 14 n ) = log 9 ( − 45 + n 2 )

ln ( x 2 − 10 ) + ln ( 9 ) = ln ( 10 ) ln ( x 2 − 10 ) + ln ( 9 ) = ln ( 10 )

For the following exercises, solve each equation for x . x .

log ( x + 12 ) = log ( x ) + log ( 12 ) log ( x + 12 ) = log ( x ) + log ( 12 )

ln ( x ) + ln ( x − 3 ) = ln ( 7 x ) ln ( x ) + ln ( x − 3 ) = ln ( 7 x )

log 2 ( 7 x + 6 ) = 3 log 2 ( 7 x + 6 ) = 3

ln ( 7 ) + ln ( 2 − 4 x 2 ) = ln ( 14 ) ln ( 7 ) + ln ( 2 − 4 x 2 ) = ln ( 14 )

log 8 ( x + 6 ) − log 8 ( x ) = log 8 ( 58 ) log 8 ( x + 6 ) − log 8 ( x ) = log 8 ( 58 )

ln ( 3 ) − ln ( 3 − 3 x ) = ln ( 4 ) ln ( 3 ) − ln ( 3 − 3 x ) = ln ( 4 )

log 3 ( 3 x ) − log 3 ( 6 ) = log 3 ( 77 ) log 3 ( 3 x ) − log 3 ( 6 ) = log 3 ( 77 )

For the following exercises, solve the equation for x , x , if there is a solution . Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

log 9 ( x ) − 5 = −4 log 9 ( x ) − 5 = −4

log 3 ( x ) + 3 = 2 log 3 ( x ) + 3 = 2

ln ( 3 x ) = 2 ln ( 3 x ) = 2

ln ( x − 5 ) = 1 ln ( x − 5 ) = 1

log ( 4 ) + log ( − 5 x ) = 2 log ( 4 ) + log ( − 5 x ) = 2

− 7 + log 3 ( 4 − x ) = −6 − 7 + log 3 ( 4 − x ) = −6

ln ( 4 x − 10 ) − 6 = − 5 ln ( 4 x − 10 ) − 6 = − 5

log ( 4 − 2 x ) = log ( − 4 x ) log ( 4 − 2 x ) = log ( − 4 x )

log 11 ( − 2 x 2 − 7 x ) = log 11 ( x − 2 ) log 11 ( − 2 x 2 − 7 x ) = log 11 ( x − 2 )

ln ( 2 x + 9 ) = ln ( − 5 x ) ln ( 2 x + 9 ) = ln ( − 5 x )

log 9 ( 3 − x ) = log 9 ( 4 x − 8 ) log 9 ( 3 − x ) = log 9 ( 4 x − 8 )

log ( x 2 + 13 ) = log ( 7 x + 3 ) log ( x 2 + 13 ) = log ( 7 x + 3 )

3 log 2 ( 10 ) − log ( x − 9 ) = log ( 44 ) 3 log 2 ( 10 ) − log ( x − 9 ) = log ( 44 )

ln ( x ) − ln ( x + 3 ) = ln ( 6 ) ln ( x ) − ln ( x + 3 ) = ln ( 6 )

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

An account with an initial deposit of $6,500 $6,500 earns 7.25 % 7.25 % annual interest, compounded continuously. How much will the account be worth after 20 years?

The formula for measuring sound intensity in decibels D D is defined by the equation D = 10 log ( I I 0 ) , D = 10 log ( I I 0 ) , where I I is the intensity of the sound in watts per square meter and I 0 = 10 − 12 I 0 = 10 − 12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.3 ⋅ 10 2 8.3 ⋅ 10 2 watts per square meter?

The population of a small town is modeled by the equation P = 1650 e 0.5 t P = 1650 e 0.5 t where t t is measured in years. In approximately how many years will the town’s population reach 20,000? 20,000?

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate the variable to 3 decimal places.

1000 ( 1.03 ) t = 5000 1000 ( 1.03 ) t = 5000 using the common log.

e 5 x = 17 e 5 x = 17 using the natural log

3 ( 1.04 ) 3 t = 8 3 ( 1.04 ) 3 t = 8 using the common log

3 4 x − 5 = 38 3 4 x − 5 = 38 using the common log

50 e − 0.12 t = 10 50 e − 0.12 t = 10 using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

7 e 3 x − 5 + 7.9 = 47 7 e 3 x − 5 + 7.9 = 47

ln ( 3 ) + ln ( 4.4 x + 6.8 ) = 2 ln ( 3 ) + ln ( 4.4 x + 6.8 ) = 2

log ( − 0.7 x − 9 ) = 1 + 5 log ( 5 ) log ( − 0.7 x − 9 ) = 1 + 5 log ( 5 )

Atmospheric pressure P P in pounds per square inch is represented by the formula P = 14.7 e − 0.21 x , P = 14.7 e − 0.21 x , where x x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 8.369 pounds per square inch? ( Hint : there are 5280 feet in a mile)

The magnitude M of an earthquake is represented by the equation M = 2 3 log ( E E 0 ) M = 2 3 log ( E E 0 ) where E E is the amount of energy released by the earthquake in joules and E 0 = 10 4.4 E 0 = 10 4.4 is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing 1.4 ⋅ 10 13 1.4 ⋅ 10 13 joules of energy?

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that b log b x = x . b log b x = x .

Recall the formula for continually compounding interest, y = A e k t . y = A e k t . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t t such that t t is equal to a single logarithm.

Recall the compound interest formula A = a ( 1 + r k ) k t . A = a ( 1 + r k ) k t . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t . t .

Newton’s Law of Cooling states that the temperature T T of an object at any time t can be described by the equation T = T s + ( T 0 − T s ) e − k t , T = T s + ( T 0 − T s ) e − k t , where T s T s is the temperature of the surrounding environment, T 0 T 0 is the initial temperature of the object, and k k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t t such that t t is equal to a single logarithm.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

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/6-6-exponential-and-logarithmic-equations

© Jan 9, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Working with Exponents and Logarithms

What is an exponent, what is a logarithm.

A Logarithm goes the other way.

It asks the question "what exponent produced this?":

And answers it like this:

In that example:

  • The Exponent takes 2 and 3 and gives 8 (2, used 3 times in a multiplication, makes 8)
  • The Logarithm takes 2 and 8 and gives 3 (2 makes 8 when used 3 times in a multiplication)

A Logarithm says how many of one number to multiply to get another number

So a logarithm actually gives us the exponent as its answer :

Working Together

Exponents and Logarithms work well together because they "undo" each other (so long as the base "a" is the same):

They are " Inverse Functions "

Doing one, then the other, gets us back to where we started:

It is too bad they are written so differently ... it makes things look strange. So it may help to think of a x as "up" and log a (x) as "down":

Anyway, the important thing is that:

The Logarithmic Function is "undone" by the Exponential Function.

(and vice versa)

Like in this example:

Example, what is x in log 3 (x) = 5

We want to "undo" the log 3 so we can get "x ="

Example: Calculate y in y = log 4 ( 1 4 )

Now a simple trick: 1 4 = 4 -1

Properties of Logarithms

One of the powerful things about Logarithms is that they can turn multiply into add .

log a ( m × n ) = log a m + log a n

"the log of multiplication is the sum of the logs"

Why is that true? See Footnote .

Using that property and the Laws of Exponents we get these useful properties:

Remember: the base "a" is always the same!

book of logarithms

And there were books full of Logarithm tables to help.

Let us have some fun using the properties:

Example: Simplify log a ( (x 2 +1) 4 √x )

That is as far as we can simplify it ... we can't do anything with log a (x 2 +1)

Answer: 4 log a (x 2 +1) + ½ log a (x)

Note: there is no rule for handling log a (m+n) or log a (m−n)

We can also apply the logarithm rules "backwards" to combine logarithms:

Example: Turn this into one logarithm: log a (5) + log a (x) − log a (2)

Answer: log a (5x/2)

The Natural Logarithm and Natural Exponential Functions

When the base is Euler's Number e = 2.718281828459... we get:

  • The Natural Logarithm log e (x) which is more commonly written ln(x)
  • The Natural Exponential Function e x

And the same idea that one can "undo" the other is still true:

ln(e x ) = x

e (ln x) = x

And here are their graphs:

They are the same curve with x-axis and y-axis flipped .

Which is another thing showing us they are inverse functions.

Always try to use Natural Logarithms and the Natural Exponential Function whenever possible.

The Common Logarithm

When the base is 10 we get:

  • The Common Logarithm log 10 (x) , which is sometimes written as log(x)

Engineers love to use it, but it is not used much in mathematics.

Example: Calculate log 10 100

Well, 10 × 10 = 100, so when 10 is used 2 times in a multiplication we get 100:

log 10 100 = 2

Likewise log 10 1,000 = 3, log 10 10,000 = 4, and so on.

Example: Calculate log 10 369

OK, best to use my calculator's "log" button:

log 10 369 = 2.567...

Changing the Base

What if we want to change the base of a logarithm?

Easy! Just use this formula:

"x goes up, a goes down"

1 log b a works as a "conversion factor" from one base to any other base.

Another useful property is:

log a x = 1 / log x a

See how "x" and "a" swap positions?

Example: Calculate 1 / log 8 2

1 / log 8 2 = log 2 8

And 2 × 2 × 2 = 8, so when 2 is used 3 times in a multiplication we get 8:

1 / log 8 2 = log 2 8 = 3

And we use the Natural Logarithm so often it is worth remembering this:

log a x = ln x / ln a

Example: Calculate log 4 22

What does this answer mean? It means that 4 with an exponent of 2.23 equals 22. So we can check that answer:

Check: 4 2.23 = 22.01 (close enough!)

Here is another example:

Example: Calculate log 5 125

We can use the "ln" function on the calculator:

Is it exactly 3? We should not trust a calculator as there could be rounding errors, but in this case we can check that 5 3 = 5 × 5 × 5 = 125 exactly , so:

Real World Usage

Here are some uses for Logarithms in the real world:

Earthquakes

The magnitude of an earthquake is a Logarithmic scale.

The famous "Richter Scale" uses this formula:

M = log 10 A + B

Where A is the amplitude (in mm) measured by the Seismograph and B is a distance correction factor

Nowadays there are more complicated formulas, but they still use a logarithmic scale.

Loudness is measured in Decibels (dB for short):

Loudness in dB = 10 log 10 (p × 10 12 )

where p is the sound pressure.

Acidic or Alkaline

Acidity (or Alkalinity) is measured in pH:

pH = −log 10 [H + ]

where H + is the molar concentration of dissolved hydrogen ions. Note: in chemistry [ ] means molar concentration (moles per liter).

More Examples

Example: solve 2 log 8 x = log 8 16.

But ... but ... but ... we can't have a log of a negative number!

So the −4 case is not defined.

Check: use a calculator to see if this is the right answer ... also try the "−4" case.

Example: Solve e − w = e 2w+6

Answer: w = − 2

Check: e -(−2) = e 2 and e 2(−2)+6 = e 2

Footnote : Why does log(m × n) = log(m) + log(n) ?

It is one of those clever things we do in mathematics which can be described as "we can't do it here, so let's go over there , do it, then come back".

solving exponential logarithmic functions

  • HW Guidelines
  • Study Skills Quiz
  • Find Local Tutors
  • Demo MathHelp.com
  • Join MathHelp.com

Select a Course Below

  • ACCUPLACER Math
  • Math Placement Test
  • PRAXIS Math
  • + more tests
  • 5th Grade Math
  • 6th Grade Math
  • Pre-Algebra
  • College Pre-Algebra
  • Introductory Algebra
  • Intermediate Algebra
  • College Algebra

Solving Exponential Equations with Logarithms

From the Definition With Logarithms With Calculators

Most exponential equations do not solve neatly; there will be no way to convert the bases to being the same, such as the conversion of 4 and 8 into powers of 2 . In solving these more-complicated equations, you will have to use logarithms.

Taking logarithms will allow us to take advantage of the log rule that says that powers inside a log can be moved out in front as multipliers. By taking the log of an exponential, we can then move the variable (being in the exponent that's now inside a log) out in front, as a multiplier on the log. In other words, the log rule will let us move the variable back down onto the ground, where we can get our hands on it.

For instance:

Content Continues Below

MathHelp.com

Change of Base Formula on MathHelp.com

Change of Base Formula

Solve 2 x = 30

If this equation had asked me to "Solve 2 x = 32 ", then finding the solution would have been easy, because I could have converted the 32 to 2 5 , set the exponents equal, and solved for " x = 5 ". But, unlike 32 , 30 is not a power of 2 so I can't set powers equal to each other. I need some other method of getting at the x , because I can't solve with the equation with the variable floating up there above the 2 ; I need it back down on the ground where it belongs, where I can get at it. And I'll have to use logarithms to bring that variable down.

When dealing with equations, I can do whatever I like to the equation, as long as I do the same thing to both sides. And, to solve an equation, I have to get the variable by itself on one side of the "equals" sign; to isolate the variable, I have to "undo" whatever has been done to the variable.

In this case, the variable x has been put in the exponent. The backwards (technically, the " inverse ") of exponentials are logarithms , so I'll need to undo the exponent by taking the log of both sides of the equation. This is useful to me because of the log rule that says that exponents inside a log can be turned into multipliers in front of the log:

log b ( m n ) = n · log b ( m )

When I take the log of both sides of an equation, I can use any log I like (base- 10 log, base- 2 log, natural log, etc), but some are sometimes more useful than others. Since the base in the equation " 2 x = 30 " is " 2 ", I might try using a base- 2 log:

log 2 (2 x ) = log 2 (30)

Any log of the log's base returns a value of 1 , so log 2 (2) = 1 . Then:

x · log 2 (2) = log 2 (30)

x (1) = log 2 (30)

x = log 2 (30)

But we can't evaluate this expression in our calculators as it stands. First, we'd need to apply the change-of-base formula to convert the expression into something in a base that our calculators can understand; namely, the natural log or the common log. That conversion looks like this:

Reminder: The " ln " is the abbreviation for "logarithmus naturalis", the Latin version of what became "natural log" in English. The abbreviation is pronounced "ell-enn" and written with a lower-case "L" followed by a lower-case "N". There is no "I" ("eye") in the function name!

What would happen if I just used the natural log, instead of a base-two log, in the first place? The process would have been exactly the same, and the eventual answer would have been equivalent.

ln(2 x ) = ln(30)

x · ln(2) = ln(30)

Either way, I get the same answer, but taking natural log in the first place was simpler and shorter.

Note: I could have used the common (base- 10 ) log instead of the natural (that is, the base- e ) log, and still come up with the same value (when evaluated in the calculator).

Since science uses the natural log so much, and since it is one of the two logs that calculators can evaluate, I tend to take the natural log of both sides when solving exponential equations. This is not (generally) required, but is often more useful than other options.

Solve 5 x = 212 . Give your answer in exact form and as a decimal approximation to three places.

Since 212 is not a power of 5 , then I will have to use logs to solve this equation. I could take base- 5 log of each side, solve, and then apply the change-of-base formula, but I think I'd rather just use the natural log in the first place:

ln(5 x ) = ln(212)

x · ln(5) = ln(212)

...or about 3.328 , rounded to three decimal places.

Solve 10 2 x = 52

Since 52 is not a power of 10 , I will have to use logs to solve this. In this particular instance, since the base is 10 and since base- 10 logs can be done on the calculator, I will use the common log instead of the natural log to solve this particular equation:

10 2 x = 52

log(10 2 x ) = log(52)

2 x · log (10) = log(52)

2 x (1) = log(52)

2 x = log(52)

...or about 0.858 , rounded to three decimal places.

Solve 3(2 x +4 ) = 350

Before I can start looking at the exponential, I first have to get rid of the 3 , so I'll divide that off to get:

...or about 2.866 , rounded to three decimal places.

Note: You could also solve the above by using exponent rules to break apart the power on the 2 :

2 x +4 = (2 x )(2 4 ) = (2 x )(16)

Then divide through by the 16 and simplify to get:

Then take the log of each side. You'll get an answer in the form:

When you evaluate this, you'll get the same decimal equivalent, 2.866 , in your calculator. Don't be shy about being flexible!

URL: https://www.purplemath.com/modules/simpexpo2.htm

Page 1 Page 2 Page 3

Standardized Test Prep

College math, homeschool math, share this page.

  • Terms of Use
  • About Purplemath
  • About the Author
  • Tutoring from PM
  • Advertising
  • Linking to PM
  • Site licencing

Visit Our Profiles

solving exponential logarithmic functions

Want Better Math Grades?

✅ Unlimited Solutions

✅ Step-by-Step Answers

✅ Available 24/7

➕ Free Bonuses ($1085 value!)

Chapter Contents ⊗

  • Exponential & Logarithmic Functions
  • 1. Definitions: Exponential and Logarithmic Functions
  • 2. Graphs of Exponential and Logarithmic Equations
  • 3. Logarithm Laws
  • 4. Logarithms to Base 10
  • 5. Natural Logarithms (base e)
  • Dow Jones Industrial Average
  • Calculating the value of e
  • 6. Exponential and Logarithmic Equations
  • World Population Live
  • 7. Graphs on Logarithmic and Semilogarithmic Axes
  • Interactive Log Table

Math Tutoring

Need help? Chat with a tutor anytime, 24/7 .

Chat Now »

Online Math Solver

Solve your math problem step by step!

Online Math Solver »

IntMath Forum

Get help with your math queries:

IntMath f orum »

Exponential and Logarithmic Functions

Why study exponential and logarithmic functions, a. exponential functions.

There are many quantities that grow exponentially . Some examples are population, compound interest and charge in a capacitor.

An understanding of exponential growth is essential if you want to be comfortably rich later on...

The special thing about exponential growth is that the rate of growth increases as time increases. You can see this in the graph at right. The curve gets steeper and steeper as time goes on.

We can also have exponential decay (for example, radioactive decay).

Related Sections in "Interactive Mathematics"

Exponents and Radicals , which is essential background before starting the current chapter.

Exponential form of a complex number

Differentiating the logarithmic function , Derivatives of exponential functions and Applications which shows how logarithms are used in calculus.

Integrating the exponential function , also part of calculus.

Differential equations: Electronics application with exponential decay and AIDS .

b. Logarithms

Logarithms were developed in the 17th century by the Scottish mathematician, John Napier. They were a clever method of reducing long multiplications into much simpler additions (and reducing divisions into subtractions). Young Johhny Napier had to help his dad, who was a tax collector. Johhny got sick of multiplying and dividing large numbers all day and devised logarithms to make his life easier.

The use of logarithms made trigonometry and many other fields of mathematics much simpler to calculate.

When calculus was developed later in the century, logarithms became central to many solutions. Today, logarithms are still important in many fields of science and engineering, even though we use calculators for most simple calculations.

You can see some applications in the "Related Sections" panel at right.

In this Chapter

  • Dow Jones Industrial Average (application)
  • World Population Live (application)

We begin the chapter with definitions of exponential and logarithmic functions »

Problem Solver

AI Math Calculator Reviews

This tool combines the power of mathematical computation engine that excels at solving mathematical formulas with the power of GPT large language models to parse and generate natural language. This creates math problem solver thats more accurate than ChatGPT, more flexible than a calculator, and faster answers than a human tutor. Learn More.

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.

Email Address Sign Up

Exponential and Logarithmic Functions

In mathematics, we might have come across different types of functions such as polynomial functions, even functions, odd functions, rational functions and trigonometric functions, etc. In this article, you will learn about a new classification of functions called exponential functions and logarithmic functions.

What are Exponential and Logarithmic Functions?

Exponential Function Definition:

An exponential function is a Mathematical function in the form y = f(x) = b x , where “x” is a variable and “b” is a constant which is called the base of the function such that b > 1. The most commonly used exponential function base is the transcendental number e, and the value of e is equal to 2.71828.

Using the base as “e” we can represent the exponential function as y = ex. This is called the natural exponential function. However, an exponential function with base 10 is called the common exponential function.

Learn more about exponential functions here.

Logarithmic Function Definition:

If the inverse of the exponential function exists then we can represent the logarithmic function as given below:

Suppose b > 1 is a real number such that the logarithm of a to base b is x if b x = a.

The logarithm of a to base b can be written as log b a.

Thus, log b a = x if b x = a.

In other words, mathematically, by making a base b > 1, we may recognise logarithm as a function from positive real numbers to all real numbers. This function is known as the logarithmic function and is defined by:

log b : R + → R

x → log b x = y if b y = x

If the base b = 10, then it is called a common logarithm and if b = e, then it is called the natural logarithm. Generally, the natural logarithm is denoted by ln.

Properties of Exponential and Logarithmic Functions

Some of the prominent features of the exponential functions are listed below:

  • The domain of the exponential function is the set of all real numbers, i.e. R.
  • The range of the exponential function is the set of all positive real numbers.
  • The point (0, 1) is always on the graph of the given exponential function since it supports the fact that b0 = 1 for any real number b > 1.
  • The exponential function is ever increasing; i.e., as we move from left to right, the graph rises above.
  • For the large set of negative values of x, the exponential function is very close to 0; for example, the graph approaches the x-axis but never meets it.

These can be observed from the graph of an exponential function given below:

Exponential and logarithmic functions 1

Some of the essential considerations on the logarithm function to any base b > 1 are listed below:

  • It is not possible to derive a meaningful definition of the logarithm for non-positive numbers, i.e. for negative numbers. So, the domain of the log function is the set of positive real numbers, i.e. R + .
  • The range of the log function is the set of all real numbers.
  • The point (1, 0) is always on the graph of the log function.
  • The log function is ever-increasing, i.e., as we move from left to right the graph rises above.
  • For the value of x quite near to zero, the value of log x can be made lesser than any given real number. That means, in quadrant IV, the graph approaches the y-axis but never meets it.

Exponential and Logarithmic functions 2

In the above graph of y = e x and y = ln x, we observe that the two curves are the mirror images of each other reflected over the line y = x.

Rules of Exponential and Logarithmic Functions

Below are the rules of exponential functions and logarithmic functions.

Exponential and Logarithmic Functions derivatives

The derivatives of exponential and logarithmic functions formulas are given below.

The derivative of e x with respect to x is written as:

The derivative of log x with respect to x is written as:

Using these two formulas, we can derive other formulas by applying exponential and logarithmic rules.

  • The derivative of eax with respect to x is: \(\begin{array}{l}\frac{d}{dx}(e^{ax})=ae^{ax}\end{array} \)
  • The nth derivative of eax with respect to x is: \(\begin{array}{l}\frac{d^n}{dx}(e^{ax})=a^ne^{ax}\end{array} \)

Similarly, we can derive multiple formulas for the derivative of exponential functions.

Exponential and Logarithmic Functions Examples

Find the derivative of log (log x), x > 1, with respect to x.

Using the formula d/dx (log x) = 1/x,

Therefore, (d/dx) [log (log x)] = 1/(x log x).

Simplify: y = 13 5 /13 3

y = 13 5 /13 3

Using the quotient rule,

y = 13 5 /13 3 = 13 (5-3)

Evaluate the derivative of the function \(\begin{array}{l}y=e^{x^4}\end{array} \) .

Given function is:

Differentiating with respect to x,

For more interesting concepts of mathematics, download BYJU’S – The Learning App today!

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

solving exponential logarithmic functions

  • Share Share

Solver Title

Practice

Generating PDF...

  • Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Mean, Median & Mode
  • Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
  • Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
  • Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
  • Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
  • Linear Algebra Matrices Vectors
  • Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
  • Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
  • Physics Mechanics
  • Chemistry Chemical Reactions Chemical Properties
  • Finance Simple Interest Compound Interest Present Value Future Value
  • Economics Point of Diminishing Return
  • Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time
  • Pre Algebra
  • One-Step Addition
  • One-Step Subtraction
  • One-Step Multiplication
  • One-Step Division
  • One-Step Decimals
  • Two-Step Integers
  • Two-Step Add/Subtract
  • Two-Step Multiply/Divide
  • Two-Step Fractions
  • Two-Step Decimals
  • Multi-Step Integers
  • Multi-Step with Parentheses
  • Multi-Step Rational
  • Multi-Step Fractions
  • Multi-Step Decimals
  • Solve by Factoring
  • Completing the Square
  • Quadratic Formula
  • Biquadratic
  • Logarithmic
  • Exponential
  • Rational Roots
  • Floor/Ceiling
  • Equation Given Roots
  • Newton Raphson
  • Substitution
  • Elimination
  • Cramer's Rule
  • Gaussian Elimination
  • System of Inequalities
  • Perfect Squares
  • Difference of Squares
  • Difference of Cubes
  • Sum of Cubes
  • Polynomials
  • Distributive Property
  • FOIL method
  • Perfect Cubes
  • Binomial Expansion
  • Negative Rule
  • Product Rule
  • Quotient Rule
  • Expand Power Rule
  • Fraction Exponent
  • Exponent Rules
  • Exponential Form
  • Logarithmic Form
  • Absolute Value
  • Rational Number
  • Powers of i
  • Partial Fractions
  • Is Polynomial
  • Leading Coefficient
  • Leading Term
  • Standard Form
  • Complete the Square
  • Synthetic Division
  • Linear Factors
  • Rationalize Denominator
  • Rationalize Numerator
  • Identify Type
  • Convergence
  • Interval Notation
  • Pi (Product) Notation
  • Boolean Algebra
  • Truth Table
  • Mutual Exclusive
  • Cardinality
  • Caretesian Product
  • Age Problems
  • Distance Problems
  • Cost Problems
  • Investment Problems
  • Number Problems
  • Percent Problems
  • Addition/Subtraction
  • Multiplication/Division
  • Dice Problems
  • Coin Problems
  • Card Problems
  • Pre Calculus
  • Linear Algebra
  • Trigonometry
  • Conversions

Click to reveal more operations

Most Used Actions

Number line.

  • \log _2(x+1)=\log _3(27)
  • \ln (x+2)-\ln (x+1)=1
  • \ln (x)+\ln (x-1)=\ln (3x+12)
  • 4+\log _3(7x)=10
  • \ln (10)-\ln (7-x)=\ln (x)
  • \log _2(x^2-6x)=3+\log _2(1-x)
  • How do you calculate logarithmic equations?
  • To solve a logarithmic equations use the esxponents rules to isolate logarithmic expressions with the same base. Set the arguments equal to each other, solve the equation and check your answer.
  • What is logarithm equation?
  • A logarithmic equation is an equation that involves the logarithm of an expression containing a varaible.
  • What are the 3 types of logarithms?
  • The three types of logarithms are common logarithms (base 10), natural logarithms (base e), and logarithms with an arbitrary base.
  • Is log10 and log the same?
  • When there's no base on the log it means the common logarithm which is log base 10.
  • What is the inverse of log in math?
  • The inverse of a log function is an exponantial.

logarithmic-equation-calculator

  • High School Math Solutions – Exponential Equation Calculator Solving exponential equations is pretty straightforward; there are basically two techniques: <ul> If the exponents... Read More

Please add a message.

Message received. Thanks for the feedback.

logo white

  • Mathematicians
  • Math Lessons
  • Square Roots
  • Math Calculators
  • Solving Logarithmic Functions – Explanation & Examples

JUMP TO TOPIC

Properties of logarithmic functions

Comparison of exponential function and logarithmic function, practice questions, solving logarithmic functions – explanation & examples.

Solving Log Function Title

Logarithms and exponents are two topics in mathematics that are closely related. Therefore it is useful we take a brief review of exponents.

An exponent is a form of writing the repeated multiplication of a number by itself. An exponential function is of the form f (x) = b y , where b > 0 < x and b ≠ 1. The quantity x is the number, b is the base, and y is the exponent or power.

For example , 32 = 2 × 2 × 2 × 2 × 2 = 2 2 .

Solving Log Function Exp and Log

On the other hand, the logarithmic function is defined as the inverse function of exponentiation. Consider again the exponential function f(x) = b y , where b > 0 < x and b ≠ 1. We can represent this function in logarithmic form as:

y = log b x

Then the logarithmic function is given by;

f(x) = log b x = y, where b is the base, y is the exponent, and x is the argument.

The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in mathematics because they enable us to perform calculations with very large numbers.

How to Solve Logarithmic Functions?

To solve the logarithmic functions, it is important to use exponential functions in the given expression. The natural log or ln is the inverse of e . That means one can undo the other one i.e.

ln (e x ) = x

To solve an equation with logarithm(s), it is important to know their properties.

Properties of logarithmic functions are simply the rules for simplifying logarithms when the inputs are in the form of division, multiplication, or exponents of logarithmic values.

Some of the properties are listed below.

  • Product rule

The product rule of logarithm states the logarithm of the product of two numbers having a common base is equal to the sum of individual logarithms.

⟹ log a  (p q) = log a  p + log a  q.

  • Quotient rule

The quotient rule of logarithms states that the logarithm of the two numbers’ ratio with the same bases is equal to the difference of each logarithm.

⟹ log a  (p/q) = log a  p – log a q

The power rule of logarithm states that the logarithm of a number with a rational exponent is equal to the product of the exponent and its logarithm.

⟹ log a  (p q ) = q log a p

  • Change of Base rule

⟹ log a p = log x p ⋅ log a x

⟹ log q p = log x p / log x q

  • Zero Exponent Rule

Solving Log Function Properties

Other properties of logarithmic functions include:

  • The bases of an exponential function and its equivalent logarithmic function are equal.
  • The logarithms of a positive number to the base of the same number are equal to 1.

log a  a = 1

  • Logarithms of 1 to any base are 0.

log a  1 = 0

  • Log a 0 is undefined
  • Logarithms of negative numbers are undefined.
  • The base of logarithms can never be negative or 1.
  • A logarithmic function with base 10is called a common logarithm. Always assume a base of 10 when solving with logarithmic functions without a small subscript for the base.

Whenever you see logarithms in the equation, you always think of how to undo the logarithm to solve the equation. For that, you use an exponential function . Both of these functions are interchangeable.

The following table tells the way of writing and interchanging the exponential functions and logarithmic functions . The third column tells about how to read both the logarithmic functions.

Let’s use these properties to solve a couple of problems involving logarithmic functions.

Rewrite exponential function 7 2 = 49 to its equivalent logarithmic function.

Given 7 2 = 64.

Here, the base = 7, exponent = 2 and the argument = 49. Therefore, 7 2 = 64 in logarithmic function is;

⟹ log 7 49 = 2

Write the logarithmic equivalent of 5 3 = 125.

exponent = 3;

and argument = 125

5 3 = 125 ⟹ log 5 125 =3

Solve for x in log  3  x = 2

log  3  x = 2 3 2  = x ⟹ x = 9

If 2 log x = 4 log 3, then find the value of ‘x’.

2 log x = 4 log 3

Divide each side by 2.

log x = (4 log 3) / 2

log x = 2 log 3

log x = log 3 2

log x = log 9

Find the logarithm of 1024 to the base 2.

1024 = 2 10

log 2 1024 = 10

Find the value of x in log 2 ( x ) = 4

Rewrite the logarithmic function log 2 ( x ) = 4 to exponential form.

Solve for x in the following logarithmic function log 2 (x – 1) = 5.

Solution Rewrite the logarithm in exponential form as;

log 2 (x – 1) = 5 ⟹ x – 1 = 2 5

Now, solve for x in the algebraic equation. ⟹ x – 1 = 32 x = 33

Find the value of x in log x 900 = 2.

Write the logarithm in exponential form as;

Find the square root of both sides of the equation to get;

x = -30 and 30

But since, the base of logarithms can never be negative or 1, therefore, the correct answer is 30.

Solve for x given, log x = log 2 + log 5

Using the product rule Log b  (m n) = log b  m + log b  n we get;

⟹ log 2 + log 5 = log (2 * 5) = Log   (10).

Therefore, x = 10.

Solve log  x  (4x – 3) = 2

Rewrite the logarithm in exponential form to get;

x 2  = 4x – 3

Now, solve the quadratic equation. x 2  = 4x – 3 x 2  – 4x + 3 = 0 (x -1) (x – 3) = 0

Since the base of a logarithm can never be 1, then the only solution is 3.

Previous Lesson  |  Main Page | Next Lesson

IMAGES

  1. Exponential Logarithmic Equations

    solving exponential logarithmic functions

  2. 29. Solving an exponential equation using logarithms

    solving exponential logarithmic functions

  3. Exponential Logarithms

    solving exponential logarithmic functions

  4. Solving Exponential Equations Using Logarithms

    solving exponential logarithmic functions

  5. Solving Logarithmic Equations

    solving exponential logarithmic functions

  6. Writing Logarithmic Equations In Exponential Form

    solving exponential logarithmic functions

VIDEO

  1. Solving Logarithmic and exponential equations 1

  2. Exponential and Logarithmic Functions Part 2

  3. 6 6 Solving Exponential and Logarithmic Equations

  4. EXPONENTIAL & LOGARITHMIC FUNCTIONS PART 1

  5. 11X1 T13 03 exponential & logarithmic functions 2023

  6. Exponential & Logarithmic Functions- Introductory Mathematical Analysis A UJ (First year BCom)

COMMENTS

  1. 4.7: Exponential and Logarithmic Equations

    Solve the resulting equation, S = T, for the unknown. Example 4.7.1: Solving an Exponential Equation with a Common Base. Solve 2x − 1 = 22x − 4. Solution. 2x − 1 = 22x − 4 The common base is 2 x − 1 = 2x − 4 By the one-to-one property the exponents must be equal x = 3 Solve for x. Exercise 4.7.1. Solve 52x = 53x + 2.

  2. Solving exponential equations using logarithms

    The exact solution is x = log 2 ( 48) . Since 48 is not a rational power of 2 , we must use the change of base rule and our calculators to evaluate the logarithm. This is shown below. x = log 2 ( 48) = log ( 48) log ( 2) Change of base rule ≈ 5.585 Evaluate using calculator The approximate solution, rounded to the nearest thousandth, is x ≈ 5.585 .

  3. Exponential & logarithmic functions

    This topic covers: - Radicals & rational exponents - Graphs & end behavior of exponential functions - Manipulating exponential expressions using exponent properties - Exponential growth & decay - Modeling with exponential functions - Solving exponential equations - Logarithm properties - Solving logarithmic equations - Graphing logarithmic func...

  4. Solving Exponential Equations Using Logarithms

    Steps to Solve Exponential Equations using Logarithms 1) Keep the exponential expression by itself on one side of the equation. 2) Get the logarithms of both sides of the equation. You can use any bases for logs. 3) Solve for the variable. Keep the answer exact or give decimal approximations.

  5. 6.6 Exponential and Logarithmic Equations

    Use logarithms to solve exponential equations. Use the definition of a logarithm to solve logarithmic equations. Use the one-to-one property of logarithms to solve logarithmic equations. Solve applied problems involving exponential and logarithmic equations. Figure 1 Wild rabbits in Australia.

  6. Solving exponential equations using logarithms: base-10

    Solving exponential equations using logarithms Solve exponential equations using logarithms: base-10 and base-e Solving exponential equations using logarithms: base-2 Solve exponential equations using logarithms: base-2 and other bases Math > Algebra 2 > Logarithms > Solving exponential equations with logarithms

  7. Calculus I

    In this section we'll take a look at solving equations with exponential functions or logarithms in them. We'll start with equations that involve exponential functions. The main property that we'll need for these equations is, logbbx = x log b b x = x Example 1 Solve 7 +15e1−3z = 10 7 + 15 e 1 − 3 z = 10 . Show Solution

  8. Solving logarithmic equations

    0:00 / 4:13 Solving logarithmic equations | Exponential and logarithmic functions | Algebra II | Khan Academy Fundraiser Khan Academy 8.26M subscribers Subscribe Subscribed 4.1K 1.3M...

  9. Working with Exponents and Logarithms

    The Logarithmic Function is "undone" by the Exponential Function. (and vice versa) Like in this example: Example, what is x in log3(x) = 5 We want to "undo" the log 3 so we can get "x =" Start with: log3 (x) = 5 Use the Exponential Function on both sides: 3log3(x) = 35 And we know that 3log3(x) = x, so: x = 35 Answer: x = 243 And also:

  10. Solving Exponential Equations with Logarithms

    Solving Exponential Equations with Logarithms From the Definition With Logarithms With Calculators Purplemath Most exponential equations do not solve neatly; there will be no way to convert the bases to being the same, such as the conversion of 4 and 8 into powers of 2. In solving these more-complicated equations, you will have to use logarithms.

  11. Exponential and Logarithmic Functions

    a. Exponential Functions. If we invest $1000 at 8% p.a., it grows to just under $5000 after 20 years. There are many quantities that grow exponentially. Some examples are population, compound interest and charge in a capacitor. An understanding of exponential growth is essential if you want to be comfortably rich later on...

  12. Logarithms

    Quiz Unit test About this unit Logarithms are the inverses of exponents. They allow us to solve challenging exponential equations, and they are a good excuse to dive deeper into the relationship between a function and its inverse. Introduction to logarithms Learn Intro to logarithms Intro to Logarithms Evaluating logarithms (advanced)

  13. Exponential and Logarithmic Functions

    Exponential Function Definition: An exponential function is a Mathematical function in the form y = f (x) = b x, where "x" is a variable and "b" is a constant which is called the base of the function such that b > 1. The most commonly used exponential function base is the transcendental number e, and the value of e is equal to 2.71828.

  14. Calculus I

    Section 1.9 : Exponential And Logarithm Equations. For problems 1 - 12 find all the solutions to the given equation. If there is no solution to the equation clearly explain why. 12−4e7+3x = 7 12 − 4 e 7 + 3 x = 7 Solution. 1 = 10−3ez2−2z 1 = 10 − 3 e z 2 − 2 z Solution. 2t−te6t−1 = 0 2 t − t e 6 t − 1 = 0 Solution.

  15. Exponential Equation Calculator

    How do you solve exponential equations? To solve an exponential equation start by isolating the exponential expression on one side of the equation. Then, take the logarithm of both sides of the equation to convert the exponential equation into a logarithmic equation.

  16. Intro to Logarithms (article)

    2 4 = 16 log 2 ( 16) = 4 Both equations describe the same relationship between the numbers 2 , 4 , and 16 , where 2 is the base and 4 is the exponent. The difference is that while the exponential form isolates the power, 16 , the logarithmic form isolates the exponent, 4 .

  17. Logarithmic Equation Calculator

    To solve a logarithmic equations use the esxponents rules to isolate logarithmic expressions with the same base. Set the arguments equal to each other, solve the equation and check your answer. What is logarithm equation? A logarithmic equation is an equation that involves the logarithm of an expression containing a varaible.

  18. Solving Logarithmic Functions

    A logarithmic function with base 10is called a common logarithm. Always assume a base of 10 when solving with logarithmic functions without a small subscript for the base. Comparison of exponential function and logarithmic function. Whenever you see logarithms in the equation, you always think of how to undo the logarithm to solve the equation.