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College Algebra

Section 15.2 Inverses of Logs and Exponentials

Chapter 4Section 15.1

Example 15.2.1.

Suppose \(f(x)=3\cdot 5^{x+2}\text{.}\) We want to find a formula for \(f^{-1}(x)\text{.}\) Recall that to find the inverse of a function, we start by writing \(y=f(x)\text{,}\) then swap \(x\) and \(y\text{,}\) then solve for \(y\text{.}\)
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Our equation starts as
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\begin{equation*} y=3\cdot 5^{x+2} \end{equation*}
When we swap \(x\) and \(y\text{,}\) we get
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\begin{equation*} x=3\cdot 5^{y+2} \end{equation*}
Now we need to solve this equation for \(y\) using what we know about solving exponential equations.
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\begin{align*} x\amp=3\cdot 5^{y+2}\\ \frac{x}{\color{red}{3}}\amp=\frac{3\cdot 5^{y+2}}{\color{red}{3}}\\ \frac{x}{3}\amp= 5^{y+2}\\ {\color{red}{\log_5}}\left(\frac{x}{3}\right)\amp= {\color{red}{\log_5}}\left(5^{y+2}\right)\\ \log_5\left(\frac{x}{3}\right)\amp= y+2\\ \log_5\left(\frac{x}{3}\right){\color{red}{-2}}\amp= y+2{\color{red}{-2}}\\ \log_5\left(\frac{x}{3}\right)-2\amp= y \end{align*}
Therefore, our answer is that \(f^{-1}(x)=\log_5\left(\frac{x}{3}\right)-2\text{.}\)
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Note that we could have used any base on the log for the previous example. If we used a different log (such as natural log), our answer would look a little different, but the function would be the same. If you used a different log and want to see if your answer is the same as our answer, try putting both into a graphing calculator, such as Desmos. If the graphs perfectly overlap, then you can be confident your answer is the same as ours!
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Checkpoint 15.2.2.

Suppose \(f(x)=2\cdot 7^{5x+4}-3\text{.}\) Find the formula for \(f^{-1}(x)\text{.}\)
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Answer.
\(f^{-1}(x)=\dfrac{\log_7\left(\frac{x+3}{2}\right)-4}{5}\)
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Solution.
Recall that to find the inverse of a function, we start by writing \(y=f(x)\text{,}\) then swap \(x\) and \(y\text{,}\) then solve for \(y\text{.}\)
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Our equation starts as
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\begin{equation*} y=2\cdot 7^{5x+4}-3 \end{equation*}
When we swap \(x\) and \(y\text{,}\) we get
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\begin{equation*} x=2\cdot 7^{5y+4}-3 \end{equation*}
Now we need to solve this equation for \(y\) using what we know about solving exponential equations.
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\begin{align*} x\amp=2\cdot 7^{5y+4}-3\\ x{\color{red}{+3}}\amp=2\cdot 7^{5y+4}-3{\color{red}{+3}}\\ x+3\amp=2\cdot 7^{5y+4}\\ \frac{x+3}{\color{red}{2}}\amp=\frac{2\cdot 7^{5y+4}}{\color{red}{2}}\\ \frac{x+3}{2}\amp= 7^{5y+4}\\ {\color{red}{\log_7}}\left(\frac{x+3}{2}\right)\amp= {\color{red}{\log_7}}\left(7^{5y+4}\right)\\ \log_7\left(\frac{x+3}{2}\right)\amp= 5y+4\\ \log_7\left(\frac{x+3}{2}\right){\color{red}{-4}}\amp= 5y+4{\color{red}{-4}}\\ \log_7\left(\frac{x+3}{2}\right)-4\amp= 5y\\ \dfrac{\log_7\left(\frac{x+3}{2}\right)-4}{\color{red}{5}}\amp= \dfrac{5y}{\color{red}{5}}\\ \dfrac{\log_7\left(\frac{x+3}{2}\right)-4}{5}\amp= y \end{align*}
Therefore, our answer is that \(f^{-1}(x)=\dfrac{\log_7\left(\frac{x+3}{2}\right)-4}{5}\text{.}\)
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Example 15.2.3.

Suppose \(g(x)=2\log_3(x-7)+4\text{.}\) We want to find the formula for \(g^{-1}(x)\text{.}\) We will follow the same strategy we always use for finding inverses. Our equation begins as
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\begin{equation*} y=2\log_3(x-7)+4 \end{equation*}
When we swap \(x\) and \(y\text{,}\) we get
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\begin{equation*} x=2\log_3(y-7)+4 \end{equation*}
Now we will solve for \(y\) using all the stragies we know for solving log equations.
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\begin{align*} x\amp=2\log_3(y-7)+4\\ x{\color{red}{-4}}\amp=2\log_3(y-7)+4{\color{red}{-4}}\\ x-4\amp=2\log_3(y-7)\\ \frac{x-4}{\color{red}{2}}\amp=\frac{2\log_3(y-7)}{\color{red}{2}}\\ \frac{x-4}{2} \amp= \log_3(y-7)\\ {\color{red}{3}}^{(x-4)/2} \amp= {\color{red}{3}}^{\log_3(y-7)}\\ 3^{(x-4)/2} \amp= y-7\\ 3^{(x-4)/2} {\color{red}{+7}}\amp= y-7{\color{red}{+7}}\\ 3^{(x-4)/2} +7\amp= y \end{align*}
Therefore, we have that \(g^{-1}(x)=3^{\frac{x-4}{2}} +7\text{.}\)
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Checkpoint 15.2.4.

Suppose \(h(x)=\log_6(2x+7)-1\text{.}\) Find \(h^{-1}(x)\text{.}\)
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Answer.
\(h^{-1}(x)=\frac{6^{x+1} -7}{2}\)
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Solution.
We will follow the same strategy we always use for finding inverses. Our equation begins as
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\begin{equation*} y=\log_6(2x+7)-1 \end{equation*}
When we swap \(x\) and \(y\text{,}\) we get
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\begin{equation*} x=\log_6(2y+7)-1 \end{equation*}
Now we will solve for \(y\) using all the stragies we know for solving log equations.
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\begin{align*} x\amp=\log_6(2y+7)-1\\ x{\color{red}{+1}}\amp=\log_6(2y+7)-1{\color{red}{+1}}\\ x+1\amp=\log_6(2y+7)\\ {\color{red}{6}}^{x+1} \amp= {\color{red}{6}}^{\log_6(2y+7)}\\ 6^{x+1} \amp= 2y+7\\ 6^{x+1} {\color{red}{-7}}\amp= 2y+7{\color{red}{-7}}\\ 6^{x+1} -7\amp= 2y\\ \frac{6^{x+1} -7}{\color{red}{2}}\amp= \frac{2y}{\color{red}{2}}\\ \frac{6^{x+1} -7}{2}\amp= y \end{align*}
Therefore, we have that \(h^{-1}(x)=\frac{6^{x+1} -7}{2}\text{.}\)
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