Lesson Objectives
• Learn how to find the Inverse of an Exponential Function
• Learn how to find the Inverse of a Logarithmic Function

## How to Find the Inverse of an Exponential or Logarithmic Function

In this lesson, we want to learn how to find the inverse of an exponential or logarithmic function. Before we get into this topic, we want to review two special properties for exponential and logarithmic expressions and how they relate to function composition. If two functions, let's say f and g are inverses, then: $$f(g(x))=x$$ For every x in the domain of g.
and $$g(f(x))=x$$ For every x in the domain of f.
Now, let's think about what happens when we use function composition with an exponential and logarithmic function. $$f(x)=a^x, a > 0, a ≠ 1$$ $$g(x)=\text{log}_a(x), a > 0, a ≠ 1, x > 0$$ $$f(g(x))=a^{\text{log}_{a}(x)}=x$$ $$g(f(x))=\text{log}_a(a^x)=x$$ We have previously seen these as special properties of logarithmic and exponential expressions:
If a > 0 and a ≠ 1: $$a^{\text{log}_a(x)}=x, x > 0$$ $$\text{log}_a(a^x)=x$$ These special properties will be extremely helpful when trying to find the inverse of an exponential or logarithmic function. Let's look at a few examples.
Example #1: Find the inverse of each. $$f(x)=\text{log}_3(2^x)$$ Step 1) Replace f(x) with y: $$y=\text{log}_3(2^x)$$ Step 2) Interchange x and y: $$x=\text{log}_3(2^y)$$ Step 3) Solve for y: $$3^x=2^y$$ $$\text{log}_{2}(3^x)=\text{log}_2(2^y)$$ $$\text{log}_{2}(3^x)=y$$ Step 4) Replace y with f-1(x): $$f^{-1}(x)=\text{log}_2(3^x)$$ Note: There are many different ways to write an answer for this type of problem. It's always best to check your answer with a graph like we did above. A quick check on Desmos will allow you to tell if the function and its inverse are reflections across the line y = x.
Let's solve the first example using a different approach to show an alternative answer. $$f(x)=\text{log}_3(2^x)$$ Step 1) Replace f(x) with y: $$y=\text{log}_3(2^x)$$ Step 2) Interchange x and y: $$x=\text{log}_3(2^y)$$ Step 3) Solve for y: $$x=y \cdot \text{log}_3(2)$$ $$y=\frac{x}{\text{log}_3(2)}$$ Step 4) Replace y with f-1(x): $$f^{-1}(x)=\frac{x}{\text{log}_3(2)}$$ Either answer will work just fine. Let's look at another example.
Example #2: Find the inverse of each. $$f(x)=\text{log}_{4}(x^5 - 7)$$ Step 1) Replace f(x) with y: $$y=\text{log}_{4}(x^5 - 7)$$ Step 2) Interchange x and y: $$x=\text{log}_{4}(y^5 - 7)$$ Step 3) Solve for y: $$4^{x}=y^5 - 7$$ $$y^5=4^{x}+ 7$$ $$y=\sqrt{4^{x}+ 7}$$ Step 4) Replace y with f-1(x): $$f^{-1}(x)=\sqrt{4^{x}+ 7}$$ Example #3: Find the inverse of each. $$f(x)=\frac{2^{x + 3}+ 1}{2^{x}}$$ Step 1) Replace f(x) with y: $$y=\frac{2^{x + 3}+ 1}{2^{x}}$$ Step 2) Interchange x and y: $$x=\frac{2^{y + 3}+ 1}{2^{y}}$$ Step 3) Solve for y: $$x=\frac{2^{y + 3}+ 1}{2^{y}}$$ $$x \cdot 2^{y}=2^{y + 3}+ 1$$ $$x \cdot 2^{y}- 2^{y + 3}=1$$ Use the rules of exponents: $$a^{b + c}=a^b \cdot a^c$$ $$x \cdot 2^{y}- 2^{y + 3}=1$$ $$x \cdot 2^{y}- 2^y \cdot 2^{3}=1$$ $$x \cdot 2^{y}- 8 \cdot 2^y=1$$ Factor out 2y: $$2^{y}(x - 8)=1$$ Divide both sides by 2y: $$x - 8=\frac{1}{2^y}$$ Use the rules of exponents: $$1^{y}=1$$ $$x - 8=\frac{1^y}{2^y}$$ $$x - 8=\left(\frac{1}{2}\right)^y$$ $$\text{log}_{\frac{1}{2}}\left(\frac{1}{2}\right)^y=\text{log}_{\frac{1}{2}}(x - 8)$$ $$y=\text{log}_{\frac{1}{2}}(x - 8)$$ Step 4) Replace y with f-1(x): $$f^{-1}(x)=\text{log}_{\frac{1}{2}}(x - 8)$$ #### Skills Check:

Example #1

Find the inverse. $$\large{f(x)=\frac{5^{x}}{3}}$$

A
$$f^{-1}(x)=\text{log}_{5}\left(\frac{x}{3}\right)$$
B
$$f^{-1}(x)=\text{log}_{5}(3x)$$
C
$$f^{-1}(x)=\text{log}_{\frac{1}{5}}(4^x)$$
D
$$f^{-1}(x)=\text{log}_{5}(x - 3)$$
E
$$f^{-1}(x)=\text{log}_{5}(3x - 5)$$

Example #2

Find the inverse. $$\large{f(x)=-2\text{log}_3(x^3) - 7}$$

A
$$f^{-1}(x)=\large{\frac{6^{\frac{x + 3}{6}}}{-3}}$$
B
$$f^{-1}(x)=3^{x - 1}+ 4$$
C
$$f^{-1}(x)=\large{\left(3^{\frac{-x - 9}{2}}\right)^{\frac{1}{2}}}$$
D
$$f^{-1}(x)=\large{\left(3^{\frac{-x - 7}{2}}\right)^{\frac{1}{3}}}$$
E
$$f^{-1}(x)=3^{x + 2}+ 7$$

Example #3

Find the inverse. $$f(x)=\frac{8 \cdot 3^x + 1}{3^x}$$

A
$$f^{-1}(x)=\text{log}_{5}(2x)$$
B
$$f^{-1}(x)=\text{log}_{5}(x + 7)$$
C
$$f^{-1}(x)=\text{log}_{3}(x - 3)$$
D
$$f^{-1}(x)=\text{log}_{\frac{1}{3}}(x - 8)$$
E
$$f^{-1}(x)=\text{log}_4(x - 1)$$         