AP®MathsCollege BoardPrecalculusRevision NotesExponential & Logarithmic FunctionsExponential & Logarithmic Equations & InequalitiesInverses of Exponential & Logarithmic Functions

Inverses of Exponential & Logarithmic Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Updated on 13 April 2026

Inverses of exponential functions

How do I find the inverse of a transformed exponential function?

A transformed exponential function has the general form
- $f (x) = a b^{(x + h)} + k$
This is a combination of
- additive transformations
  - horizontal shift $h$
  - vertical shift $k$
- and a vertical dilation
  - factor $a$
- applied to the base exponential function $b^{x}$
To find the inverse, think of the function as a sequence of operations applied to the input $x$
- and then reverse each operation in the opposite order
- Make sure you are familiar with the process for finding an inverse function

In general this will proceed as follows (using $y = a b^{(x + h)} + k$ )
- Start with the equation
  - $y = a b^{(x + h)} + k$
- Subtract $k$
  - $y - k = a b^{(x + h)}$
- Divide by $a$
  - $\frac{y - k}{a} = b^{(x + h)}$
- Take $\log_{b}$ of both sides (this 'cancels' the exponential on the right)
  - $\log_{b} (\frac{y - k}{a}) = x + h$
- Subtract $h$
  - $x = \log_{b} (\frac{y - k}{a}) - h$
- Swap $x$ and $y$
  - $f^{- 1} (x) = \log_{b} (\frac{x - k}{a}) - h$

The inverse of a transformed exponential function is always a transformed logarithmic function

How does this work in practice?

E.g. to find the inverse of $f (x) = 3 \cdot 2^{(x + 1)} - 5$
Think about the operations that transform $x$ into $f (x)$
- For $f (x) = 3 \cdot 2^{(x + 1)} - 5$ , the operations in order are
  - Add 1
    - $x \to x + 1$
  - Raise 2 to the power of the result
    - $x + 1 \to 2^{(x + 1)}$
  - Multiply by 3
    - $2^{(x + 1)} \to 3 \cdot 2^{(x + 1)}$
  - Subtract 5
    - $3 \cdot 2^{(x + 1)} \to 3 \cdot 2^{(x + 1)} - 5$

To find the inverse, apply the inverse operations in the opposite order to $y = 3 \cdot 2^{(x + 1)} - 5$
- Add 5 (reverses 'subtract 5')
  - $y + 5 = 3 \cdot 2^{(x + 1)}$
- Divide by 3 (reverses 'multiply by 3')
  - $\frac{y + 5}{3} = 2^{(x + 1)}$
- Take $\log_{2}$ (reverses 'raise 2 to the power')
  - and $\log_{2}$ cancels $2^{□}$ , so
  - $\log_{2} (\frac{y + 5}{3}) = x + 1$
- Subtract 1 (reverses 'add 1')
  - $\log_{2} (\frac{y + 5}{3}) - 1 = x$
- Swap $x$ and $y$
  - $f^{- 1} (x) = \log_{2} (\frac{x + 5}{3}) - 1$

What is the domain of the inverse?

The domain of the inverse function equals the range of the original function
- and vice versa
E.g. for $f (x) = a b^{(x + h)} + k$
- If $a > 0$
  - the range of $f$ is $(k, \infty)$
    - so the domain of $f^{- 1}$ is $(k, \infty)$
- If $a < 0$
  - the range of $f$ is $(- \infty, k)$ , so the domain of $f^{- 1}$ is $(- \infty, k)$
This makes sense because the logarithmic expression in the inverse $f^{- 1} (x) = \log_{b} (\frac{x - k}{a}) - h$ requires a positive argument

Inverses of logarithmic functions

How do I find the inverse of a transformed logarithmic function?

A transformed logarithmic function has the general form $f (x) = a \log_{b} (x + h) + k$
This is a combination of
- additive transformations
  - horizontal shift $h$
  - vertical shift $k$
- and a vertical dilation
  - factor $a$
- applied to the base logarithmic function $\log_{b} x$
As with finding the inverse of an exponential, find the inverse by reversing the operations in the opposite order

In general (using $y = a \log_{b} (x + h) + k$ )
- Start with the equation
  - $y = a \log_{b} (x + h) + k$
- Subtract $k$
  - $y - k = a \log_{b} (x + h)$
- Divide by $a$
  - $\frac{y - k}{a} = \log_{b} (x + h)$
- Convert to exponential form ( $b^{□}$ cancels $\log_{b}$ on the right-hand side)
  - $b^{(y - k) / a} = x + h$
- Subtract $h$
  - $b^{(y - k) / a} - h = x$
- Swap $x$ and $y$ :
  - $f^{- 1} (x) = b^{(x - k) / a} - h$

The inverse of a transformed logarithmic function is always a transformed exponential function

E.g. to find the inverse of $f (x) = 2 \log_{3} (x - 4) + 1$
- follow the steps given above
  - $y = 2 \log_{3} (x - 4) + 1$
  - $y - 1 = 2 \log_{3} (x - 4)$
  - $\frac{y - 1}{2} = \log_{3} (x - 4)$
  - $3^{(y - 1) / 2} = x - 4$
  - $x = 3^{(y - 1) / 2} + 4$
  - $f^{- 1} (x) = 3^{(x - 1) / 2} + 4$

What is the domain of the inverse?

The domain of the inverse function equals the range of the original function
- and vice versa
For $f (x) = a \log_{b} (x + h) + k$
- The range of a logarithmic function (in general form) is all real numbers
- So the domain of the inverse (which is an exponential function) is also all real numbers
This is consistent with what you know about exponential functions having a domain of all real numbers

Examiner Tips and Tricks

A quick way to check your answer is to verify that $f^{- 1} (f (x)) = x$ for a simple value of $x$ .

Worked Example

The function $f$ is given by

$f (x) = 5 e^{(x - 2)} + 3$

Find $f^{- 1} (x)$ . Be sure to indicate the domain of $f^{- 1}$ .

Answer:

Write

$y = 5 e^{(x - 2)} + 3$

Subtract 3

$y - 3 = 5 e^{(x - 2)}$

Divide by 5

$\frac{y - 3}{5} = e^{(x - 2)}$

Take the natural logarithm of both sides

$\ln$ cancels $e^{□}$ on the right-hand side

$\ln (\frac{y - 3}{5}) = x - 2$

Add 2

$x = \ln (\frac{y - 3}{5}) + 2$

Swap $x$ and $y$

$f^{- 1} (x) = \ln (\frac{x - 3}{5}) + 2$

You can check this using a simple value of $x$ , say $x = 2$

First find $f (2)$

$f (2) = 5 e^{0} + 3 = 5 + 3 = 8$

Then substitute that value into $f^{- 1} (x)$ to see if $f^{- 1} (f (2)) = 2$

$f^{- 1} (8) = \ln (\frac{8 - 3}{5}) + 2 = \ln (1) + 2 = 0 + 2 = 2$ ✓

To find the domain, note that the argument of the logarithm must be positive

$\frac{x - 3}{5} > 0 \Rightarrow x > 3$

So the answer is

$f^{- 1} (x) = \ln (\frac{x - 3}{5}) + 2$ , with domain $x > 3$

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Inverses of Exponential & Logarithmic Functions (College Board AP® Precalculus): Revision Note

Inverses of exponential functions

How do I find the inverse of a transformed exponential function?

How does this work in practice?

What is the domain of the inverse?

Inverses of logarithmic functions

How do I find the inverse of a transformed logarithmic function?

What is the domain of the inverse?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B