### About Inverse of an Exponential / Logarithmic Function:

In some cases, we will need to find the inverse of an exponential function: f(x) = ax, where a is greater than zero and a is not equal to 1. Additionally, we may need to find the inverse of a logarithmic function: f(x) = loga(x), where a is greater than zero, a is not equal to 1, and x is greater than zero.

Test Objectives
• Demonstrate the ability to find the inverse of an exponential function
• Demonstrate the ability to find the inverse of a logarithmic function
Inverse of an Exponential / Logarithmic Function Practice Test:

#1:

Instructions: find the inverse.

$$a)\hspace{.2em}f(x)=\text{log}_{3}(x^3 + 2) + 7$$

$$b)\hspace{.2em}f(x)=7\text{log}_{2}(x - 3) - 3$$

#2:

Instructions: find the inverse.

$$a)\hspace{.2em}f(x)=\text{log}_{3}(-3x - 6) - 4$$

$$b)\hspace{.2em}f(x)=\left(\frac{4^x + 4}{4}\right)^{\frac{1}{4}}$$ For this problem, you need to consider the range. This will become the domain in the inverse. $$\text{Range}:\{y | y > 1\}$$

#3:

Instructions: find the inverse.

$$a)\hspace{.2em}f(x)=\left(\frac{4^x - 2}{-4}\right)^{\frac{1}{4}}$$ For this problem, you need to consider the range. This will become the domain in the inverse. $$\text{Range}:\left\{y | 0 ≤ y < \frac{\sqrt[4]{8}}{2}\right\}$$

$$b)\hspace{.2em}f(x)=\frac{-1}{2^{x + 2}}$$

#4:

Instructions: find the inverse.

$$a)\hspace{.2em}f(x)=4^x + 4$$

$$b)\hspace{.2em}f(x)=\frac{6^x}{2}$$

#5:

Instructions: find the inverse.

$$a)\hspace{.2em}f(x)=\frac{1}{\sqrt[4]{3^x}}$$ For this problem, you need to consider the range. This will become the domain in the inverse. $$\text{Range}:\{y | y > 0\}$$

$$b)\hspace{.2em}f(x)=\frac{4 \cdot 3^x + 1}{3^x}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\sqrt[3]{3^{x - 7}- 2}$$

Graph Key:

$$f^{-1}(x)=\sqrt[3]{3^{x - 7}- 2}$$

$$y=x$$

$$f(x)=\text{log}_{3}(x^3 + 2) + 7$$

$$b)\hspace{.2em}f^{-1}(x)=\large{2^{\frac{x + 3}{7}}}+ 3$$

Graph Key:

$$f^{-1}(x)=\large{2^{\frac{x + 3}{7}}}+ 3$$

$$y=x$$

$$f(x)=7\text{log}_{2}(x - 3) - 3$$

#2:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=-3^{x + 3}- 2$$

Graph Key:

$$f^{-1}(x)=-3^{x + 3}- 2$$

$$y=x$$

$$f(x)=\text{log}_{3}(-3x - 6) - 4$$

$$b)\hspace{.2em}f^{-1}(x)=\text{log}_{4}(4x^4 - 4), x > 1$$

Graph Key:

$$f^{-1}(x)=\text{log}_{4}(4x^4 - 4), x > 1$$

$$y=x$$

$$f(x)=\left(\frac{4^x + 4}{4}\right)^{\frac{1}{4}}$$

#3:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\text{log}_{4}(-4x^4 + 2), 0 ≤ x < \frac{\sqrt[4]{8}}{2}$$

Graph Key:

$$f^{-1}(x)=\text{log}_{4}(-4x^4 + 2), 0 ≤ x < \frac{\sqrt[4]{8}}{2}$$

$$y=x$$

$$f(x)=\left(\frac{4^x - 2}{-4}\right)^{\frac{1}{4}}$$

$$b)\hspace{.2em}f^{-1}(x)=\text{log}_{\frac{1}{2}}(-4x)$$

Graph Key:

$$f^{-1}(x)=\text{log}_{\frac{1}{2}}(-4x)$$

$$y=x$$

$$f(x)=\frac{-1}{2^{x + 2}}$$

#4:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\text{log}_{4}(x - 4)$$

Graph Key:

$$f^{-1}(x)=\text{log}_{4}(x - 4)$$

$$y=x$$

$$f(x)=4^x + 4$$

$$b)\hspace{.2em}f^{-1}(x)=\text{log}_{6}(2x)$$

Graph Key:

$$f^{-1}(x)=\text{log}_{6}(2x)$$

$$y=x$$

$$f(x)=\frac{6^x}{2}$$

#5:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\text{log}_{\frac{1}{3}}(x^4), x > 0$$

Graph Key:

$$f^{-1}(x)=\text{log}_{\frac{1}{3}}(x^4), x > 0$$

$$y=x$$

$$f(x)=\frac{1}{\sqrt[4]{3^x}}$$

$$b)\hspace{.2em}f^{-1}(x)=\text{log}_{\frac{1}{3}}(x - 4)$$

Graph Key:

$$f^{-1}(x)=\text{log}_{\frac{1}{3}}(x - 4)$$

$$y=x$$

$$f(x)=\frac{4 \cdot 3^x + 1}{3^x}$$