Precalculus · Logarithms and exponentials

How to Solve a Logarithmic Equation

Problem

log⁡2x=5\log_2 x = 5

Answer

x=32x = 32

A logarithmic equation becomes a power equation once you use the definition. Raise the base to the value on the other side to recover the argument.

Step-by-step solution

  1. Rewrite the logarithm as a power.

    log⁡bx=y⇒by=x\log_b x = y \Rightarrow b^y = x

    Why: A logarithm is an exponent, so the definition reverses the statement.

  2. Apply the definition with base 2 and exponent 5.

    25=x2^5 = x

    Why: The base of the log becomes the base of the power.

  3. Evaluate the power.

    x=25=32x = 2^5 = 32

    Why: Two to the fifth power is 32, and 32 is positive, so it is in the domain.

Common mistakes

  • Multiplying 2 by 5 and writing x = 10. The log definition uses a power, not a product.
  • Swapping the base and the argument.
  • Forgetting that x must be positive. Logs only accept positive inputs.

Practice problems

Use the same method. Work each problem on paper, then open the answer to check.

  1. log⁡3x=2\log_3 x = 2

    Show answer

    x=9x = 9

  2. log⁡2(x+1)=3\log_2(x + 1) = 3

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    x=7x = 7

  3. log⁡5x=0\log_5 x = 0

    Show answer

    x=1x = 1

  4. log⁡4x=2\log_4 x = 2

    Show answer

    x=16x = 16

  5. log⁡10x=2\log_{10} x = 2

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    x=100x = 100