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

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 13 April 2026

Logarithmic equations & inequalities

How do I solve a logarithmic equation?

A logarithmic equation is an equation that involves one or more logarithmic expressions containing the variable
- E.g. $\log_{3} x = 5$ or $\ln (x + 2) + \ln (x - 1) = \ln 10$
The key tool for solving logarithmic equations is the inverse relationship between logarithmic and exponential functions
- $\log_{b} x = c \Leftrightarrow x = b^{c}$
  - this includes the case with base $e$ :
    - $\ln x = c \Leftrightarrow x = e^{c}$
- This allows you to convert a logarithmic expression into an exponential one

Method 1: Converting directly to exponential form
- If the equation has a single logarithmic expression equal to a constant
  - then you can convert using the definition of a logarithm
- E.g. to solve $\log_{5} x = 3$
  - Rewrite in exponential form: $x = 5^{3} = 125$
- E.g. to solve $\ln (x + 3) = 2$
  - Rewrite: $x + 3 = e^{2}$
  - Solve: $x = e^{2} - 3$

Method 2: Using the one-to-one property
- If two logarithmic expressions with the same base are equal
  - their arguments must be equal
- I.e. if $\log_{b} A = \log_{b} B$ , then $A = B$
  - provided $A > 0$ and $B > 0$
- E.g. to solve $\log_{10} (x^{2} - 3) = \log_{10} (2 x)$
  - Set the arguments equal: $x^{2} - 3 = 2 x$
  - Rearrange: $x^{2} - 2 x - 3 = 0$
  - Factor: $(x - 3) (x + 1) = 0$
  - So $x = 3$ or $x = - 1$
- After solving in this way, check for extraneous solutions (i.e. invalid solutions)
  - This is discussed below

Method 3: Combining logarithms first, then solving
- If the equation has multiple logarithmic terms
  - you can use the properties of logarithms to combine them into a single logarithm, then solve
- The key properties used are:
  - Product property: $\log_{b} (x y) = \log_{b} x + \log_{b} y$
  - Quotient property: $\log_{b} (\frac{x}{y}) = \log_{b} x - \log_{b} y$
  - Power property: $\log_{b} (x^{n}) = n \log_{b} x$
- E.g. to solve $\log_{10} (x - 1) + \log_{10} (x + 3) = \log_{10} (x + 9)$
  - Use the product property on the left-hand side
    - $\log_{10} ((x - 1) (x + 3)) = \log_{10} (x + 9)$
  - Apply the one-to-one property
    - $(x - 1) (x + 3) = x + 9$
  - Expand and simplify
    - $x^{2} + 2 x - 3 = x + 9$
    - giving $x^{2} + x - 12 = 0$
  - Factor
    - $(x + 4) (x - 3) = 0$
  - So $x = - 4$ or $x = 3$
- After solving in this way, check for extraneous solutions (i.e. invalid solutions)
  - This is discussed below

Examiner Tips and Tricks

When combining logarithms using the product or quotient property, remember that you can only do this when the logarithms have the same base!

Examiner Tips and Tricks

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

E.g. giving an answer in terms of $e$ , rather than as a decimal approximation

On calculator questions, decimal approximations are often expected.

Make sure to give answers correct to three decimal places

Why do logarithmic equations produce extraneous solutions?

Logarithmic equations frequently produce extraneous solutions
- these are values that satisfy the algebraic equation
- but are not valid in the original logarithmic equation
This happens because logarithms are only defined for positive arguments
- I.e. $\log_{b} x$ is only defined when $x > 0$
- So the domain of any logarithmic expression restricts which values of $x$ are valid as solutions

You must always check your solutions against the domain of the original equation
- Substitute each solution back into every logarithmic expression in the original equation
- If any argument becomes zero or negative, that solution is extraneous and must be rejected

E.g. the equation $\log_{10} (x - 1) + \log_{10} (x + 3) = \log_{10} (x + 9)$ was solved above to find $x = - 4$ or $x = 3$
- For $x = 3$
  - $\log_{10} (2) + \log_{10} (6) = \log_{10} (12)$
  - all arguments (2, 6 and 12) are positive, so $x = 3$ is valid
- For $x = - 4$
  - $\log_{10} (- 5) + \log_{10} (- 1) = \log_{10} (5)$
  - $\log_{10} (- 5)$ and $\log_{10} (- 1)$ are both undefined
  - so $x = - 4$ is extraneous
- The only valid solution is $x = 3$

Extraneous solutions can also arise from contextual limitations in applied problems
- E.g. if the variable represents a physical quantity like time or concentration, additional constraints on what values those quantities might take may apply

Examiner Tips and Tricks

Always check for extraneous solutions.

With logarithmic equations, this is even more important than with exponential equations
The domain restriction (argument must be positive) very commonly eliminates one or more algebraic solutions

What about equations involving a coefficient in front of a logarithm?

The power property of logarithms means that a coefficient in front of a logarithm can be interpreted as an exponent inside the logarithm
- $n \log_{b} x = \log_{b} (x^{n})$
Using this to solve a logarithmic equation can affect the number of solutions found
- And some of those solutions might be extraneous

E.g. consider the equation $2 \log_{3} x = 4$
One way to solve it is to divide both sides by 2 first
- This gives $\log_{3} x = 2$
  - so $x = 3^{2} = 9$
Another approach is to use the power property first
- This gives $\log_{3} (x^{2}) = 4$
  - so $x^{2} = 3^{4} = 81$
  - giving $x = 9$ or $x = - 9$
- Note that $x = - 9$ is a valid solution to $\log_{3} (x^{2}) = 4$
  - because $(- 9)^{2} = 81 > 0$
However, the two approaches give different solutions because
- $\log_{3} x$ is only defined for $x > 0$
- while $\log_{3} (x^{2})$ is defined for all $x \neq 0$
To be valid a solution must satisfy the equation as given in the question
- So in this case only $x = 9$ is a valid solution
- because the original equation contains $\log_{3} x$ , not $\log_{3} (x^{2})$

Examiner Tips and Tricks

When using the power property to combine or simplify logarithms, pay attention to whether the original equation uses $n \log_{b} x$ or $\log_{b} (x^{n})$ .

These have different domains
and this distinction has appeared on the exam

How do I solve logarithmic inequalities?

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

E.g. to solve $\ln (x - 1) > 3$
- Start by solving as an equation $\ln (x - 1) = 3$
  - rewrite in exponential form: $x - 1 = e^{3}$
- $e > 1$ so $\ln (x - 1) = \log_{e} (x - 1)$ is an increasing function
  - That means $\ln (x - 1) > 3$ for $x - 1 > e^{3}$
- Add 1 to both sides for the final solution
  - $x > 1 + e^{3}$
You must also ensure that all arguments remain positive throughout the solution
- The domain restriction (that the argument inside a logarithm must be greater than 0) acts as an additional constraint that must be combined with the inequality solution
- In the above example
  - $x > 1 + e^{3} \Rightarrow x - 1 > e^{3} > 0$
  - so the solution is valid

Examiner Tips and Tricks

On free response questions, always show your work step by step.

The scoring guidelines generally require supporting work to earn full credit

Worked Example

The function $g$ is given by

$g (x) = 3 \log_{4} x$

Solve $g (x) = 6$ for values of $x$ in the domain of $g$ . Show the work that leads to your answer.

Answer:

Set $g (x) = 6$

$3 \log_{4} x = 6$

Divide both sides by 3

$\log_{4} x = 2$

Convert to exponential form

$x = 4^{2} = 16$

Since $x = 16 > 0$ , this is in the domain of $g$

$x = 16$

Worked Example

Consider the functions $f$ and $g$ given by $f (x) = \ln (x + 2) + \ln (x - 1)$ and $g (x) = \ln (5 x - 2)$ . In the $x y$ -plane, what are all $x$ -coordinates of the points of intersection of the graphs of $f$ and $g$ ?

(A) $x = 4$ only

(B) $x = 0$ and $x = 4$

(D) $x = 0$ only

Answer:

The $x$ -coordinates of any intersection points will the solutions to the equation $f (x) = g (x)$

$\ln (x + 2) + \ln (x - 1) = \ln (5 x - 2)$

Use the product property on the left side

$\ln ((x + 2) (x - 1)) = \ln (5 x - 2)$

Apply the one-to-one property (if $\ln A = \ln B$ then $A = B$ )

$(x + 2) (x - 1) = 5 x - 2$

Expand the left side

$x^{2} + x - 2 = 5 x - 2$

Simplify

$\begin{array}{rcl} x^{2} + x - 5 x - 2 + 2 & = & 0 \\ x^{2} - 4 x & = & 0 \end{array}$

Factor and solve

$x (x - 4) = 0$

$x = 0$ or $x = 4$

Now check for extraneous solutions by verifying the domain of the original equation

Every logarithmic argument must be positive.

For $x = 4$ :
$x + 2 = 6 > 0$
$x - 1 = 3 > 0$
$5 x - 2 = 18 > 0$
All positive, so valid

For $x = 0$ :
$x + 2 = 2 > 0$
$x - 1 = - 1 < 0$
$5 x - 2 = - 2 < 0$
Two are negative, so not valid

For $x = 0$ , $\ln (x - 1)$ and $\ln (5 x - 2)$ are not defined

Therefore the only valid solution to the equation is $x = 4$

(A) $x = 4$ only

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

Logarithmic equations & inequalities

How do I solve a logarithmic equation?

Examiner Tips and Tricks

Examiner Tips and Tricks

Why do logarithmic equations produce extraneous solutions?

Examiner Tips and Tricks

What about equations involving a coefficient in front of a logarithm?

Examiner Tips and Tricks

How do I solve logarithmic inequalities?

Examiner Tips and Tricks

Worked Example

Worked Example

Unlock more, it's free!

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

Author: Roger B

Reviewer: Mark Curtis