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Solving Exponential Equations & Inequalities (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 13 April 2026

Exponential equations & inequalities

How do I solve an exponential equation?

An exponential equation is an equation where the variable appears in an exponent
- E.g. $3^{x} = 81$ or $5 e^{(x + 3)} = 40$
The general strategy for solving exponential equations involves isolating the exponential expression
- and then using logarithms (or properties of exponents) to solve for the variable

Method 1: Rewriting with a common base
- If both sides of the equation can be written as powers of the same base, you can set the exponents equal
- E.g. to solve $2^{(x + 1)} = 16$
  - Rewrite $16$ as $2^{4}$
  - So $2^{(x + 1)} = 2^{4}$
  - Therefore $x + 1 = 4$ , giving $x = 3$
- This method works well when the numbers involved are recognizable powers of a common base

Method 2: Taking logarithms of both sides
- When the equation cannot be rewritten with a common base, take the logarithm of both sides
- You can use any base of logarithm
  - but the natural logarithm ( $\ln$ ) is most common, especially with base $e$ equations
- E.g. to solve $e^{(x + 3)} = 10$
  - Take $\ln$ of both sides: $\ln (e^{(x + 3)}) = \ln 10$
  - Simplify the left side: $x + 3 = \ln 10$
    - $\ln$ cancels $e^{◻}$ because they are inverses
  - Solve: $x = - 3 + \ln 10$
- For equations with bases other than $e$ , the same approach works
- E.g. to solve $2.916 \cdot (0.7)^{x} = 2$
  - Divide both sides by 2.916: $(0.7)^{x} = \frac{2}{2.916}$
  - Take $\ln$ of both sides: $\ln (0 . 7^{x}) = \ln (\frac{2}{2.916})$
  - Use properties of logarithms: $x \ln (0.7) = \ln (\frac{2}{2.916})$
  - Divide: $x = \frac{\ln (2 / 2.916)}{\ln (0.7)} \approx 1.057$

Method 3: Using the inverse relationship $b^{x} = c \Leftrightarrow x = \log_{b} c$
- This is equivalent to taking logarithms
  - but uses the definition of a logarithm directly
- E.g. $e^{(x + 3)} = 10$ means $x + 3 = \log_{e} 10 = \ln 10$

The key property that makes all these methods work is the inverse relationship between exponential and logarithmic functions

Examiner Tips and Tricks

Remember that you can use logarithms and exponentials (with the same base) to 'cancel' each other when solving equations:

$\log_{b} (b^{x}) = x$
- This includes $\ln (e^{x}) = x$
$b^{\log_{b} x} = x$
- This includes $e^{\ln x} = x$

Examiner Tips and Tricks

On non-calculator questions, you are expected to find solutions as exact values.

Leave answers in terms of $\ln$ , $\log$ , or as simplified expressions rather than decimal approximations

On calculator questions, decimal approximations are often expected.

Make sure to give answers correct to three decimal places

What about exponential equations that involve additional terms?

Sometimes you need to isolate the exponential term before applying logarithms
- E.g. to solve $4 e^{(x - 2)} + 1 = 5$
  - Subtract 1: $4 e^{(x - 2)} = 4$
  - Divide by 4: $e^{(x - 2)} = 1$
  - Since $e^{0} = 1$ , it follows that $x - 2 = 0$
  - so $x = 2$

In general, for equations of the form $a e^{(b x + c)} + d = k$ :
- Subtract $d$ : $a e^{(b x + c)} = k - d$
- Divide by $a$ : $e^{(b x + c)} = \frac{k - d}{a}$
- Take $\ln$ : $b x + c = \ln (\frac{k - d}{a})$
- Solve for $x$ : $x = \frac{\ln (\frac{k - d}{a}) - c}{b}$

After isolating the exponential expression, always check whether the right-hand side is positive
- Since $b^{x} > 0$ for any base $b > 0$ , an equation like $e^{x} = - 3$ has no solution
- This connects to checking for extraneous solutions (see below)

What is a 'hidden quadratic' exponential equation?

Some exponential equations can be transformed into quadratic equations using a substitution
- This happens when the equation contains terms involving both $b^{2 x}$ and $b^{x}$
- Since $b^{2 x} = (b^{x})^{2}$ , substituting $y = b^{x}$ converts the equation into a quadratic in $y$

E.g. to solve $e^{2 x} - 5 e^{x} + 6 = 0$ :
- Let $y = e^{x}$ , so $e^{2 x} = y^{2}$
  - The equation becomes $y^{2} - 5 y + 6 = 0$
- Factor: $(y - 2) (y - 3) = 0$
  - So $y = 2$ or $y = 3$
- Back-substitute
  - $e^{x} = 2 \Rightarrow x = \ln 2$
  - or $e^{x} = 3 \Rightarrow x = \ln 3$

This technique also works with bases other than $e$
E.g. $4^{x} - 3 \cdot 2^{x} + 2 = 0$
- can be rewritten as $(2^{x})^{2} - 3 \cdot 2^{x} + 2 = 0$
  - because $4^{x} = {(2^{2})}^{x} = 2^{2 x} = {(2^{x})}^{2}$
- Let $y = 2^{x}$
  - $y^{2} - 3 y + 2 = 0$
  - giving $(y - 1) (y - 2) = 0$
- So $2^{x} = 1 \Rightarrow x = 0$
  - or $2^{x} = 2 \Rightarrow x = 1$

What are extraneous solutions, and why do I need to check for them?

When solving exponential equations, you should always check whether the solutions you find are valid
Solutions can be extraneous (invalid) for two reasons:
Mathematical limitations
- The range of an exponential function means that $b^{x} > 0$ for all real $x$ (when $b > 0$ )
- So if a substitution or manipulation leads to $b^{x} = negative number$
  - there is no solution from that part
- E.g. if solving a hidden quadratic gives $e^{x} = - 5$ as one of the answers
  - this must be rejected because $e^{x}$ is always positive
- This is the most common source of extraneous solutions in exponential equations

Contextual limitations
- In applied problems, the domain may be restricted
  - E.g. if $t$ represents time and $t \geq 0$ , then a negative solution for $t$ would be extraneous
- Or if a model is only valid for certain input values
  - then solutions outside that set of values should be rejected

Examiner Tips and Tricks

When solving hidden quadratic equations in a free response question, always show your substitution clearly (e.g. "Let $y = e^{x}$ ").

The exam commonly awards a point specifically for demonstrating the quadratic form

Also remember to reject any extraneous solutions and briefly explain why.

E.g. " $e^{x} = - 5$ has no solution because $e^{x} > 0$ for all $x$ "

How do I solve exponential inequalities?

Exponential inequalities are solved using similar techniques to exponential equations
A key principle here is that an exponential function is either always increasing or always decreasing
- This depends on the value of the base $b$
  - If $b > 1$ , the exponential function $b^{x}$ is increasing
  - If $0 < b < 1$ , the exponential function $b^{x}$ is decreasing
This means that if $x = p$ is the solution to $b^{x} = k$
- then for everywhere on one side of $p$
  - $b^{x} > k$ is true
- and for everywhere on the other side of $p$
  - $b^{x} < k$ is true
- $b^{x}$ being increasing or decreasing will tell you which side is which

E.g. $0 . 5^{x} > 0.125$
- Start by solving as an equation $0 . 5^{x} = 0.125$
  - take $\log_{0.5}$ of both sides: $\log_{0.5} (0 . 5^{x}) = \log_{0.5} (0.125)$
  - $\log_{0.5}$ cancels $0 . 5^{◻}$ : $x = \log_{0.5} (0.125) = 3$
- $0.5 < 1$ so $0 . 5^{x}$ is a decreasing function
  - That means $0 . 5^{x} > 0.125$ for $x < 3$
E.g. $e^{2 x} > 7$
- Start by solving as an equation $e^{2 x} = 7$
  - take $\ln$ of both sides: $\ln (e^{2 x}) = \ln 7$
  - $\ln$ cancels $e^{◻}$ : $2 x = \ln 7$
- $e > 1$ so $e^{2 x}$ is an increasing function
  - That means $e^{2 x} > 7$ for $2 x > \ln 7$
- Divide by 2 for the final solution
  - $x > \frac{\ln 7}{2}$

How can logarithms help reveal useful information about exponential expressions?

The identity $b^{x} = c^{(x \log_{c} b)}$ allows you to convert between bases for an exponential function
- This is covered in the Logarithmic Functions as Inverses of Exponential Functions study guide
This is useful when you need to compare or combine exponential expressions with different bases
- For example, $2^{x}$ can be rewritten as $e^{x \ln 2}$
- This can be useful for solving equations that mix different exponential bases

Worked Example

The function $P$ is given by $P (t) = \frac{300, 000}{1 + 0.7 e^{k t}}$ , where $k$ is a constant. If $P (3) = 200, 000$ , what is the value of $P (12)$ ?

(A) 176,471

(B) 184.532

(D) 253,761

Answer:

Method 1 (finding $k$ explicitly)

Substitute $t = 3$ and $P (3) = 200, 000$ into the equation for $P (t)$

$200000 = \frac{300000}{1 + 0.7 e^{3 k}}$

Rearrange to get $e^{3 k}$ on its own on one side of the equation

$\begin{array}{rcl} 200000 (1 + 0.7 e^{3 k}) & = & 300000 \\ 1 + 0.7 e^{3 k} & = & \frac{300000}{200000} \\ 1 + 0.7 e^{3 k} & = & 1.5 \\ 0.7 e^{3 k} & = & 1.5 - 1 \\ 0.7 e^{3 k} & = & 0.5 \\ e^{3 k} & = & \frac{0.5}{0.7} \\ e^{3 k} & = & \frac{5}{7} \end{array}$

Take the natural logarithm of both sides

$\ln$ and $e^{◻}$ are inverse functions, to they will cancel on the left-hand side

$\begin{array}{rcl} \ln (e^{3 k}) & = & \ln (\frac{5}{7}) \\ 3 k & = & \ln (\frac{5}{7}) \end{array}$

Then you can solve for $k$

$\begin{array}{rcl} k & = & \frac{\ln (\frac{5}{7})}{3} = - 0.112157412 . . . \end{array}$

Substitute that value of $k$ along with $t = 12$ into the formula for $P (t)$ to find $P (12)$

$P (12) = \frac{300000}{1 + 0.7 e^{(\begin{array}{rcl} - \end{array} \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 112157412 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \times 12)}} = 253760.789 . . .$

To the nearest integer that is 253761

(D) 253,761

Method 2 (clever shortcut that avoids having to find $k$ )

This relies on using laws of indices to realise that

$e^{12 k} = e^{3 k \times 4} = {(e^{3 k})}^{4}$

Solve as in Method 1 up to

$\begin{array}{rcl} e^{3 k} & = & \frac{5}{7} \end{array}$

Then substitute $t = 12$ into the formula for $P (t)$

$P (12) = \frac{300000}{1 + 0.7 e^{12 k}}$

and use the two results from above

$= \frac{300000}{1 + 0.7 {(e^{3 k})}^{4}} = \frac{300000}{1 + 0.7 {(\frac{5}{7})}^{4}} = 253760.7891$

(D) 253,761

Worked Example

The function $f$ is given by

$f (x) = e^{2 x} + 2 e^{x} - 15$

Find all values of $x$ in the domain of $f$ for which $f (x) = 0$ . Show the work that leads to your answer.

Answer:

Set $f (x) = 0$

$e^{2 x} + 2 e^{x} - 15 = 0$

Note that $e^{2 x} = (e^{x})^{2}$ , so this is a 'hidden' quadratic equation in $e^{x}$

Let $y = e^{x}$ :

$y^{2} + 2 y - 15 = 0$

Factor the quadratic

$(y + 5) (y - 3) = 0$

So $y = - 5$ or $y = 3$

Back-substitute $y = e^{x}$

$e^{x} = - 5$ or $e^{x} = 3$

Deal with the two possibilities

Note that $e^{x} = - 5$ is an extraneous solution

$e^{x} = - 5$ has no solution, because $e^{x} > 0$ for all real values of $x$

$e^{x} = 3$ can be solved by taking the natural logarithm of both sides

$\ln (e^{x}) = \ln 3$

And $\ln$ cancels $e^{◻}$ , so

$x = \ln 3$

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Solving Exponential Equations & Inequalities (College Board AP® Precalculus): Study Guide

Exponential equations & inequalities

How do I solve an exponential equation?

Examiner Tips and Tricks

Examiner Tips and Tricks

What about exponential equations that involve additional terms?

What is a 'hidden quadratic' exponential equation?

What are extraneous solutions, and why do I need to check for them?

Examiner Tips and Tricks

How do I solve exponential inequalities?

How can logarithms help reveal useful information about exponential expressions?

Worked Example

Worked Example

Unlock more, it's free!

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

Author: Roger B

Reviewer: Mark Curtis