Solving Logarithmic Equations Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Solving Logarithmic Equations.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Solving equations containing logarithms by converting to exponential form or using log properties to combine and simplify.

If logarithms trap the variable inside a log⁡\log, converting to exponential form releases it. The key insight is that log⁡b(stuff)=c\log_b(\text{stuff}) = c means bc=stuffb^c = \text{stuff}—just rewrite and solve.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Rewrite log⁡b(stuff)=c\log_b(\text{stuff})=c as bc=stuffb^c=\text{stuff} to free the variable from the logarithm.

Common stuck point: The procedure for solving logarithmic equations is the easy part; the trap is skipping the domain check. Asking "Is the unknown the argument of a log, so I rewrite as bc=argumentb^c=\text{argument} to release it?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Is the unknown the argument of a log, so I rewrite as bc=argumentb^c=\text{argument} to release it?

Worked Examples

Example 1

easy
Solve log⁡2(x)=5\log_2(x) = 5.

Answer

x=32x = 32

First step

1
Convert from logarithmic to exponential form: log⁡2(x)=5\log_2(x) = 5 means 25=x2^5 = x.

Full solution

  1. 2
    Calculate: x=25=32x = 2^5 = 32.
  2. 3
    Check: log⁡2(32)=log⁡2(25)=5\log_2(32) = \log_2(2^5) = 5. ✓
The fundamental relationship between logarithms and exponents is: log⁡b(a)=c\log_b(a) = c if and only if bc=ab^c = a. Converting to exponential form is the most direct way to solve simple logarithmic equations.

Example 2

medium
Solve log⁡(x+3)+log⁡(x−1)=log⁡(5)\log(x+3) + \log(x-1) = \log(5).

Example 3

easy
Solve log⁡3(9)=x\log_3(9) = x.

Example 4

medium
Solve log⁡2(x)+log⁡2(x+2)=3\log_2(x) + \log_2(x+2) = 3.

Example 5

hard
Solve ln⁡(x)+ln⁡(x−1)=ln⁡6\ln(x) + \ln(x-1) = \ln 6.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

medium
Solve ln⁡(x2−4)=ln⁡(3x)\ln(x^2 - 4) = \ln(3x).

Example 2

hard
Solve log⁡3(x+6)−log⁡3(x−2)=2\log_3(x+6) - \log_3(x-2) = 2.

Example 3

easy
Solve log⁡2x=3\log_2 x = 3.

Example 4

easy
Solve log⁡3x=2\log_3 x = 2.

Example 5

easy
Solve log⁡10x=0\log_{10} x = 0.

Example 6

easy
Solve log⁡5x=1\log_5 x = 1.

Example 7

easy
Solve log⁡4x=12\log_4 x = \frac12.

Example 8

easy
Solve log⁡2(x+1)=4\log_2(x+1) = 4.

Example 9

easy
Solve ln⁡x=0\ln x = 0.

Example 10

easy
Solve log⁡3(2x)=2\log_3(2x) = 2.

Example 11

medium
Solve log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x-2) = 3.

Example 12

medium
Solve log⁡3(x+6)−log⁡3x=2\log_3(x+6) - \log_3 x = 2.

Example 13

medium
Solve log⁡(x)+log⁡(x−3)=1\log(x) + \log(x-3) = 1 (base 10).

Example 14

medium
Solve log⁡2(x2−1)=3\log_2(x^2-1) = 3.

Example 15

medium
Solve 2log⁡3x=log⁡3162\log_3 x = \log_3 16.

Example 16

medium
Solve log⁡5(3x−2)=log⁡5(x+6)\log_5(3x-2) = \log_5(x+6).

Example 17

medium
Solve ln⁡(x)=2\ln(x) = 2, leaving the answer in terms of ee.

Example 18

challenge
Solve log⁡2x+log⁡4x=3\log_2 x + \log_4 x = 3.

Example 19

challenge
Solve (log⁡2x)2−3log⁡2x+2=0(\log_2 x)^2 - 3\log_2 x + 2 = 0.

Example 20

challenge
Solve log⁡2(x)+log⁡2(x+6)=4\log_2(x) + \log_2(x+6) = 4.

Example 21

medium
Solve log⁡6(x)+log⁡6(x−1)=1\log_6(x) + \log_6(x-1) = 1.

Example 22

medium
Solve 3log⁡2x=63\log_2 x = 6.

Example 23

easy
Solve log⁡2x=4\log_2 x = 4.

Example 24

easy
Solve log⁡5x=3\log_5 x = 3.

Example 25

easy
Solve log⁡2(x−1)=3\log_2(x-1) = 3.

Example 26

easy
Solve log⁡10(100x)=4\log_{10}(100x) = 4.

Example 27

easy
Solve log⁡2x=−1\log_2 x = -1.

Example 28

medium
Solve log⁡4(x)+log⁡4(x−6)=2\log_4(x) + \log_4(x-6) = 2.

Example 29

medium
Solve log⁡2(x+4)−log⁡2(x−1)=3\log_2(x+4) - \log_2(x-1) = 3.

Example 30

medium
Solve ln⁡(2x+1)=3\ln(2x+1) = 3.

Example 31

medium
Solve log⁡3(2x−1)=log⁡3(x+4)\log_3(2x-1) = \log_3(x+4).

Example 32

medium
Solve 2log⁡5x=log⁡5812\log_5 x = \log_5 81.

Example 33

medium
Solve log⁡10(x)−log⁡10(x−3)=1\log_{10}(x) - \log_{10}(x-3) = 1.

Example 34

medium
Solve log⁡5(x2−24)=2\log_5(x^2 - 24) = 2.

Example 35

hard
Solve log⁡3(x)+log⁡9(x)=3\log_3(x) + \log_9(x) = 3.

Example 36

hard
Solve (log⁡3x)2=log⁡3x+6(\log_3 x)^2 = \log_3 x + 6.

Example 37

hard
A loud noise has dB=10log⁡10(I/I0)\text{dB} = 10\log_{10}(I/I_0) with I0=10−12 W/m2I_0 = 10^{-12}\text{ W/m}^2. Find II if dB=85\text{dB} = 85.

Example 38

hard
Solve log⁡3(x+5)+log⁡3(x−3)=2\log_3(x+5) + \log_3(x-3) = 2.

Example 39

hard
Solve log⁡2(log⁡3x)=1\log_2(\log_3 x) = 1.

Example 40

hard
Solve log⁡x16=2\log_x 16 = 2.

Example 41

challenge
Solve xlog⁡10x=100xx^{\log_{10} x} = 100x.

Example 42

challenge
Solve log⁡2x+log⁡4x+log⁡8x=11\log_2 x + \log_4 x + \log_8 x = 11.
← Back to Solving Logarithmic Equations Formula Explained Practice Mode

Background Knowledge

These ideas may be useful before you work through the harder examples.

logarithmlogarithm properties