Inverses of exponential functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: One-to-one verification of exponential functions, definition of inverse exponential functions as logarithms, finding inverses of transformed exponentials, and graphical properties of inverse exponential functions.

You should already know: 1. How to test if a function is one-to-one via the horizontal line test or algebraic test. 2. Core properties of inverse functions (domain/range swap, reflection over $y=x$). 3. Basic algebraic manipulation of exponential expressions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Inverses of exponential functions?

An inverse of an exponential function is the function that reverses the input-output mapping of the original exponential function. All exponential functions of the form $f(x)=b^{kx+c}$ (where $b>0,b\ne 1,k\ne 0$) are strictly increasing or strictly decreasing, so they are one-to-one and have a valid inverse function over their entire domain. This inverse function is universally known as a logarithmic function, so the study of inverses of exponentials is the foundation of all logarithmic function work in AP Precalculus. Per the AP Precalculus Course and Exam Description (CED), this topic is a core foundational concept for Unit 2, which accounts for ~25% of the total AP exam score, and inverses of exponentials specifically make up roughly 10% of Unit 2 content. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections: you can expect 1-2 MCQs testing basic inverse properties, and often a part of a multi-part FRQ that asks to invert a contextual exponential model for interpretation.

2. One-to-Oneness of Exponential Functions

To confirm a function has an inverse that is also a function, it must pass the horizontal line test: no horizontal line intersects the graph of the function more than once. Algebraically, this means for any $x_1\ne x_2$, $f(x_1)\ne f(x_2)$. For exponential functions with a linear exponent of the form $f(x)=b^{kx+c}$ where $b>0,b\ne 1$ and $k\ne 0$, this property always holds. We can prove this algebraically: if we suppose $b^{k{x}_1+c}=b^{k{x}_2+c}$, we can take the natural logarithm of both sides to get $(k{x}_1+c)\ln b=(k{x}_2+c)\ln b$. Since $\ln b\ne 0$ for $b\ne 1$ and $k\ne 0$, we can simplify to get $x_1=x_2$, which confirms one-to-oneness. The only time an exponential function is not one-to-one is if its exponent is non-linear (e.g., $f(x)=2^{x^2}$, which maps $x$ and $-x$ to the same output) or its domain is restricted in a way that creates duplicate outputs.

Worked Example

Problem: Confirm whether $f(x)=5^{3-2x}$ has a valid inverse function over its entire domain of $\mathbb{R}$, and justify your answer. Solution:

1. Start with the algebraic one-to-one test: assume $f(x_1)=f(x_2)$. We get $5^{3-2x_1}=5^{3-2x_2}$.
2. Take the natural logarithm of both sides: $\ln \left(5^{3-2x_1}\right)=\ln \left(5^{3-2x_2}\right)$.
3. Apply the logarithm power rule: $(3-2x_1)\ln 5=(3-2x_2)\ln 5$.
4. Since $5>0$ and $5\ne 1$, $\ln 5\ne 0$, so we can divide both sides by $\ln 5$ to get $3-2x_1=3-2x_2$.
5. Simplify: $-2x_1=-2x_2⟹x_1=x_2$. Conclusion: Since $f(x_1)=f(x_2)$ implies $x_1=x_2$, $f(x)$ is one-to-one over $\mathbb{R}$ and has a valid inverse function over its entire domain.

Exam tip: If you are asked to justify one-to-oneness on an FRQ, always use either the horizontal line test for graphs or the algebraic test shown above; stating "exponentials are always one-to-one" without justification will not earn full credit.

3. Finding the Inverse of a Transformed Exponential Function

Once we confirm an exponential function is one-to-one, we can find its inverse (a logarithmic function) using the standard inverse function procedure: swap $x$ and $y$, then solve for $y$. By definition, the inverse of $f(x)=b^x$ is $f^{-1}(x)={\log }_{b}x$, which comes directly from the definition of inverses: if $y=b^x$, swapping gives $x=b^y$, which is equivalent to $y={\log }_{b}x$ by definition of the logarithm. For transformed exponentials of the form $y=a\cdot b^{kx+c}+d$, we follow the same steps to solve for $y$, resulting in a transformed logarithmic inverse. Remember that for any inverse, the domain of the original function becomes the range of the inverse, and the range of the original becomes the domain of the inverse. Since the range of any non-constant exponential function is $(d,\infty )$ if $a>0$ (or $(-\infty ,d)$ if $a<0$), the domain of the inverse logarithmic function will match that interval.

Worked Example

Problem: Find the inverse of $f(x)=3\cdot 2^{4x-1}+2$, state the domain of the inverse, and verify using the inverse function identity $f(f^{-1}(x))=x$. Solution:

1. Start by writing $y=3\cdot 2^{4x-1}+2$, then swap $x$ and $y$ to get $x=3\cdot 2^{4y-1}+2$.
2. Isolate the exponential term: subtract 2 from both sides, then divide by 3: $\frac{x-2}{3}=2^{4y-1}$.
3. Convert from exponential form to logarithmic form: ${\log }_2\left(\frac{x-2}{3}\right)=4y-1$.
4. Solve for $y$: add 1 to both sides, divide by 4: $y=\frac{1}{4}\left({\log }_2\left(\frac{x-2}{3}\right)+1\right)=f^{-1}(x)$.
5. Domain check: The range of the original $f(x)$ is $(2,\infty )$, so the domain of $f^{-1}(x)$ is $(2,\infty )$.
6. Verify: $f(f^{-1}(x))=3\cdot 2^{4\left(\frac{1}{4}\left({\log }_2\left(\frac{x-2}{3}\right)+1\right)\right)-1}+2=3\cdot 2^{{\log }_2\left(\frac{x-2}{3}\right)}+2=(x-2)+2=x$, which confirms the inverse is correct.

Exam tip: Always state the domain of your inverse function on FRQ questions; AP exam graders regularly deduct points for missing domain restrictions on inverses.

4. Graphical Properties of Inverse Exponential Functions

All inverse functions have graphs that are reflections of the original function's graph over the line $y=x$, and inverses of exponentials are no exception: the graph of $y={\log }_{b}x$ is the exact reflection of $y=b^x$ over $y=x$. This property lets us quickly identify key points, intercepts, asymptotes, and end behavior of the inverse without finding its full equation. For example, the original exponential $y=b^x$ has a y-intercept at $(0,1)$, a horizontal asymptote at $y=0$, domain $\mathbb{R}$, and range $(0,\infty )$. Reflecting over $y=x$ swaps the coordinates of all key points, so the inverse $y={\log }_{b}x$ has an x-intercept at $(1,0)$, a vertical asymptote at $x=0$, domain $(0,\infty )$, and range $\mathbb{R}$, which matches the domain/range swap rule for inverses. For transformed exponentials, this reflection rule still holds, and we can swap asymptote type and point coordinates directly.

Worked Example

Problem: The original function $f(x)=4^{x+2}-3$ has a horizontal asymptote at $y=-3$ and passes through the point $(-1,1)$. State the asymptote and a point that lies on the graph of $f^{-1}(x)$, without finding the equation of the inverse. Solution:

1. Recall that all points on the graph of $f$ are of the form $(a,f(a))$, so corresponding points on $f^{-1}$ are of the form $(f(a),a)$.
2. The given point on $f$ is $(-1,1)$, so swapping coordinates gives $(1,-1)$ as a point on $f^{-1}(x)$.
3. Asymptotes of inverse functions swap types: a horizontal asymptote $y=k$ on the original function becomes a vertical asymptote $x=k$ on the inverse function.
4. So the horizontal asymptote $y=-3$ on $f(x)$ becomes a vertical asymptote $x=-3$ on $f^{-1}(x)$. Final answer: $f^{-1}(x)$ has vertical asymptote $x=-3$ and passes through $(1,-1)$.

Exam tip: When asked to graph an inverse exponential, plot 2-3 key reflected points and the swapped asymptote first, then draw the curve; this avoids mistakes with end behavior.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming all functions containing an exponential term are one-to-one, e.g., claiming $f(x)=3^{x^2}$ has an inverse over all real numbers. Why: Students generalize that "all exponential functions are one-to-one" but forget this only applies to exponentials with linear exponents, not non-linear exponents that produce duplicate outputs. Correct move: Always test one-to-oneness for any exponential function with a non-linear exponent by checking if $f(x_1)=f(x_2)$ can hold for $x_1\ne x_2$.
• Wrong move: When solving for the inverse of $y=2b^{kx+c}+d$, incorrectly rearranging to $x-2=b^{kx+c}-d$ instead of isolating the exponential correctly. Why: Students rush isolating the exponential term and incorrectly subtract the leading coefficient instead of dividing. Correct move: Always isolate the exponential term step-by-step: first add/subtract the constant term, then multiply/divide by the leading coefficient to get the exponential alone on one side.
• Wrong move: Forgetting to swap the domain and range when stating the domain of the inverse, so writing the domain of the inverse exponential as $(-\infty ,\infty )$, the same as the original exponential. Why: Students confuse the domain of the original exponential with the domain of its inverse, and forget inverse functions swap domain and range. Correct move: After finding the inverse equation, write down the range of the original exponential first; that range is automatically the domain of your inverse.
• Wrong move: Reflecting the asymptote incorrectly: changing a horizontal asymptote $y=5$ on the original exponential to $y=5$ on the inverse instead of $x=5$. Why: Students remember reflection over $y=x$ but forget that horizontal lines map to vertical lines and vice versa. Correct move: For inverse reflections: any horizontal asymptote $y=k$ becomes vertical asymptote $x=k$, and any vertical asymptote $x=k$ becomes horizontal asymptote $y=k$.
• Wrong move: When verifying an inverse, computing only $f^{-1}(f(x))$ and stopping, instead of checking both compositions. Why: Students forget that the inverse identity requires both compositions to hold, especially for functions with restricted domains. Correct move: On verification questions, always confirm both compositions simplify to $x$ to earn full credit.
• Wrong move: Converting $x=b^{y+a}$ incorrectly to $y={\log }_{b}x+a$ instead of $y={\log }_{b}x-a$. Why: Students rush the algebra when pulling the exponent out of the logarithm and misapply rearrangement rules. Correct move: After getting to $x=b^{y+a}$, explicitly write ${\log }_{b}x=y+a$, then rearrange step-by-step to solve for $y$.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following gives the inverse of $f(x)=4e^{2x}-7$ and correctly states the domain of the inverse? A) $f^{-1}(x)=\frac{1}{2}\ln \left(\frac{x+7}{4}\right)$, domain $(-7,\infty )$ B) $f^{-1}(x)=\ln \left(\frac{x+7}{2}\right)$, domain $(-7,\infty )$ C) $f^{-1}(x)=\frac{1}{2}\ln \left(\frac{x-7}{4}\right)$, domain $(7,\infty )$ D) $f^{-1}(x)=2\ln \left(\frac{x+7}{4}\right)$, domain $(-\infty ,\infty )$

Worked Solution: We follow the standard inverse procedure for $f(x)=4e^{2x}-7$. First set $y=4e^{2x}-7$, then swap $x$ and $y$ to get $x=4e^{2y}-7$. Isolate the exponential term: add 7 to both sides to get $x+7=4e^{2y}$, then divide by 4 to get $\frac{x+7}{4}=e^{2y}$. Take the natural logarithm of both sides: $\ln \left(\frac{x+7}{4}\right)=2y$, so rearrange to get $y=\frac{1}{2}\ln \left(\frac{x+7}{4}\right)$. The original function $f(x)$ has range $(-7,\infty )$, so the domain of the inverse is also $(-7,\infty )$. This matches option A. Correct answer: A.

Question 2 (Free Response)

Let $f(x)=2\cdot 3^{x+1}-18$. (a) Show that $f(x)$ is one-to-one over its domain of all real numbers, and find $f^{-1}(x)$. (b) State the domain and range of $f^{-1}(x)$. (c) Evaluate $f^{-1}(36)$, show your working.

Worked Solution: (a) To confirm one-to-oneness, assume $f(x_1)=f(x_2)$: $$2\cdot 3^{x_1+1}-18=2\cdot 3^{x_2+1}-18$$ Add 18 to both sides and divide by 2: $3^{x_1+1}=3^{x_2+1}$. Take the natural logarithm of both sides: $(x_1+1)\ln 3=(x_2+1)\ln 3$. Divide by $\ln 3\ne 0$: $x_1+1=x_2+1⟹x_1=x_2$, so $f(x)$ is one-to-one. To find the inverse: Set $y=2\cdot 3^{x+1}-18$, swap $x$ and $y$: $x=2\cdot 3^{y+1}-18$. Isolate the exponential: $\frac{x+18}{2}=3^{y+1}$, convert to logarithmic form: ${\log }_3\left(\frac{x+18}{2}\right)=y+1$, so $f^{-1}(x)={\log }_3\left(\frac{x+18}{2}\right)-1$.

(b) The domain of the original $f(x)$ is all real numbers, and its range is $(-18,\infty )$. For inverses, domain of the inverse = range of the original, and range of the inverse = domain of the original. So domain of $f^{-1}(x)$ is $(-18,\infty )$, and range is $(-\infty ,\infty )$.

(c) Substitute $x=36$ into $f^{-1}(x)$: $$f^{-1}(36)={\log }_3\left(\frac{36+18}{2}\right)-1={\log }_3(27)-1=3-1=2$$ Verification: $f(2)=2\cdot 3^3-18=54-18=36$, which confirms the result.

Question 3 (Application / Real-World Style)

A conservation biologist is modeling the population of a growing wolf pack as a function of time $t$ (in years) after reintroduction: $P(t)=8\cdot 2^{0.25t}$, where $P(t)$ is the number of wolves. Researchers want to predict how long it will take for the population to reach a targeted size, so they need an inverse function that gives time as a function of population size. Find this inverse function, and use it to calculate how many years it will take for the wolf population to reach 64 wolves.

Worked Solution: We solve for $t$ in terms of $P$ to get the inverse function:

1. Start with the original model: $P=8\cdot 2^{0.25t}$.
2. Isolate the exponential term: divide both sides by 8 to get $\frac{P}{8}=2^{0.25t}$.
3. Convert to logarithmic form: ${\log }_2\left(\frac{P}{8}\right)=0.25t$.
4. Solve for $t$: $t(P)=4{\log }_2\left(\frac{P}{8}\right)$, which is the inverse function.
5. Substitute $P=64$: $t(64)=4{\log }_2\left(\frac{64}{8}\right)=4{\log }_2(8)=4(3)=12$.

Interpretation: It will take 12 years after reintroduction for the wolf population to grow from 8 to 64 wolves.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational base for all work with logarithmic functions, which are the core of the rest of AP Precalculus Unit 2. Immediately after mastering inverses of exponentials, you will move on to properties of logarithms, solving exponential and logarithmic equations, and modeling with logarithmic functions for real-world scenarios. Without a solid understanding of how logarithms are defined as inverses of exponentials, you will not be able to correctly justify logarithm properties, solve equations, or interpret logarithmic models in context. This topic also reinforces broader AP Precalculus concepts like one-to-one functions, inverse function properties, transformations of function families, and contextual modeling of growth and decay. Follow-up topics to study next:

• Properties of logarithms
• Solving exponential and logarithmic equations
• Modeling with exponential and logarithmic functions