Solve Exponential and Logarithmic Equations

By the end of this section, you will be able to:

Solve logarithmic equations using the properties of logarithms
Solve exponential equations using logarithms
Use exponential models in applications

Before you get started, take this readiness quiz.

Solve: $x^{2} = 16 .$

If you missed this problem, review [link].
Solve: $x^{2} - 5 x + 6 = 0 .$

If you missed this problem, review [link].
Solve: $x (x + 6) = 2 x + 5 .$

If you missed this problem, review [link].

Solve Logarithmic Equations Using the Properties of Logarithms

In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations.

If our equation has two logarithms we can use a property that says that if ${log}_{a} M = {log}_{a} N$

then it is true that $M = N .$

This is the One-to-One Property of Logarithmic Equations.

One-to-One Property of Logarithmic Equations

For $M > 0, N > 0, a > 0,$

and $a \neq 1$

is any real number:

If {log}_{a} M = {log}_{a} N, then M = N .

To use this property, we must be certain that both sides of the equation are written with the same base.

Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number.

Solve: $2 {log}_{5} x = {log}_{5} 81 .$

\begin{matrix} \begin{matrix} 2 \log_{5} x & = & \log_{5} 81 \\ Use the Power Property. & \log_{5} x^{2} & = & \log_{5} 81 \\ Use the One-to-One Property, if \log_{a} M = \log_{a} N, & x^{2} & = & 81 \\ then M = N . \\ Solve using the Square Root Property. & x & = & ± 9 \end{matrix} \\ We eliminate x = −9 as we cannot take the logarithm x = 9, x = −9 \\ of a negative number. \\ Check. \\ \begin{matrix} x = 9 & 2 \log_{5} x & = & \log_{5} 81 \\ 2 \log_{5} 9 & \overset{?}{=} & \log_{5} 81 \\ \log_{5} 9^{2} & \overset{?}{=} & \log_{5} 81 \\ \log_{5} 81 & = & \log_{5} 81 ✓ \end{matrix} \end{matrix}

Solve: $2 {log}_{3} x = {log}_{3} 36$

x = 6

Solve: $3 log x = log 64$

x = 4

Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm.

Solve: ${log}_{3} x + {log}_{3} (x - 8) = 2 .$

\begin{matrix} \begin{matrix} \log_{3} x + \log_{3} (x - 8) & = & 2 \\ Use the Product Property, \log_{a} M + \log_{a} N = \log_{a} M \cdot N . & \log_{3} x (x - 8) & = & 2 \\ Rewrite in exponential form. & 3^{2} & = & x (x - 8) \\ Simplify. & 9 & = & x^{2} - 8 x \\ Subtract 9 from each side. & 0 & = & x^{2} - 8 x - 9 \\ Factor. & 0 & = & (x - 9) (x + 1) \\ Use the Zero-Product Property. & x - 9 & = & 0, x + 1 = 0 \end{matrix} \\ Solve each equation. x = 9, x = −1 \\ Check. \\ \begin{matrix} x = −1 & \log_{3} x + \log_{3} (x - 8) & = & 2 \\ \log_{3} (−1) + \log_{3} (−1 −8) & \overset{?}{=} & 2 \end{matrix} \\ We cannot take the log of a negative number. \\ \begin{matrix} x = 9 & \log_{3} x + \log_{3} (x - 8) & = & 2 \\ \log_{3} 9 + \log_{3} (9 - 8) & \overset{?}{=} & 2 \\ 2 + 0 & \overset{?}{=} & 2 \\ 2 & = & 2 ✓ \end{matrix} \end{matrix}

Solve: ${log}_{2} x + {log}_{2} (x - 2) = 3$

x = 4

Solve: ${log}_{2} x + {log}_{2} (x - 6) = 4$

x = 8

When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed.

Solve: ${log}_{4} (x + 6) - {log}_{4} (2 x + 5) = − {log}_{4} x .$

\begin{matrix} \begin{matrix} {log}_{4} (x + 6) - {log}_{4} (2 x + 5) & = & − {log}_{4} x \\ \begin{matrix} Use the Quotient Property on the left side and the Power \\ Property on the right. \end{matrix} & {log}_{4} (\frac{x + 6}{2 x + 5}) & = & {log}_{4} x^{−1} \\ Rewrite x^{−1} = \frac{1}{x} . & {log}_{4} (\frac{x + 6}{2 x + 5}) & = & {log}_{4} \frac{1}{x} \\ \begin{matrix} Use the One-to-One Property, if {log}_{a} M = {log}_{a} N, \\ then M = N . \end{matrix} & \frac{x + 6}{2 x + 5} & = & \frac{1}{x} \\ Solve the rational equation. & x (x + 6) & = & 2 x + 5 \\ Distribute. & x^{2} + 6 x & = & 2 x + 5 \\ Write in standard form. & x^{2} + 4 x - 5 & = & 0 \\ Factor. & (x + 5) (x - 1) & = & 0 \end{matrix} \\ Use the Zero-Product Property. x + 5 = 0, x - 1 = 0 \\ Solve each equation. x = −5, x = 1 \\ Check. \\ We leave the check for you. \end{matrix}

Solve: $log (x + 2) - log (4 x + 3) = − log x .$

x = 3

Solve: $log (x - 2) - log (4 x + 16) = log \frac{1}{x} .$

x = 8

Solve Exponential Equations Using Logarithms

In the section on exponential functions, we solved some equations by writing both sides of the equation with the same base. Next we wrote a new equation by setting the exponents equal.

It is not always possible or convenient to write the expressions with the same base. In that case we often take the common logarithm or natural logarithm of both sides once the exponential is isolated.

Solve $5^{x} = 11 .$