Unit 11 · Lesson 5 of 5

Solving logarithmic equations

About 25 minutes · 5 steps · one idea: undo the log, then check the inputs

You'll be able to: solve equations like \(\log_3(2x+1) = 2\) and \(\log_2 x + \log_2(x-2) = 3\), and throw out answers that make a log undefined.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first, then check.

Solve \(2^{x+3} = 16\).

\(16 = 2^4\), so \(x + 3 = 4\) and \(x = 1\).

Solve \(x^2 - x - 6 = 0\).

Two numbers that multiply to \(-6\) and add to \(-1\): \(-3\) and \(2\). So \((x - 3)(x + 2) = 0\).

Set each factor to 0: \(x = 3\) or \(x = -2\).

Step 2 of 5The idea

To solve a log equation, get rid of the log. There are two common shapes, and both end with a check.

Log = number. Rewrite in exponential form: \(\log_b(\text{stuff}) = c\) becomes \(\text{stuff} = b^c\).

Log = log (same base). Drop the logs: \(\log_b M = \log_b N\) means \(M = N\).

More than one log on a side? Condense into one log first (Lesson 0056).

Always check. Put each answer back into every log of the original equation. Each log's input must be positive. An answer that fails is an extraneous solution: cross it out.

Why can a wrong answer sneak in? Condensing \(\log_2 x + \log_2(x - 2)\) into \(\log_2\big(x(x-2)\big)\) makes the equation accept some negative \(x\) values that the original logs could never take.

Why this shows up on the placement test: log equations are on the MATH 1010 skill list, and the test likes answer choices that include the extraneous value.

Step 3 of 5Worked examples

Example 1: \(\log_3(2x + 1) = 2\)

  1. Exponential form: \(2x + 1 = 3^2\). Base 3, raised to the 2, gives the inside.
  2. \(2x + 1 = 9\), so \(2x = 8\) and \(x = 4\).
  3. Check: \(2(4) + 1 = 9\), which is positive, and \(\log_3 9 = 2\). So \(x = 4\).

Example 2: \(\log_2 x + \log_2(x - 2) = 3\)

  1. Condense: \(\log_2\big(x(x - 2)\big) = 3\). Product rule: plus becomes times.
  2. Exponential form: \(x(x - 2) = 2^3 = 8\), so \(x^2 - 2x - 8 = 0\).
  3. Factor: \((x - 4)(x + 2) = 0\), so \(x = 4\) or \(x = -2\).
  4. Check \(x = 4\): \(\log_2 4 + \log_2 2 = 2 + 1 = 3\). It works.
  5. Check \(x = -2\): \(\log_2(-2)\) is undefined. Extraneous, so cross it out. Answer: \(x = 4\) only.

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. No calculator needed. If a value is extraneous, leave it out.

Solve \(\log_2 x = 5\).

Exponential form: \(x = 2^5 = 32\). Check: 32 is positive. Answer: \(x = 32\).

Solve \(\log x = -2\).

No base written means base 10: \(x = 10^{-2} = \dfrac{1}{100}\) (or 0.01). It is positive, so it works.

Solve \(\log_5(2x + 5) = 2\).

Exponential form: \(2x + 5 = 5^2 = 25\).

\(2x = 20\), so \(x = 10\). Check: \(2(10) + 5 = 25 \gt 0\). Answer: \(x = 10\).

Solve \(\ln(x + 4) = \ln(3x - 2)\).

Log = log, so the insides are equal: \(x + 4 = 3x - 2\).

\(6 = 2x\), so \(x = 3\). Check: \(3 + 4 = 7\) and \(3(3) - 2 = 7\), both positive. Answer: \(x = 3\).

Solve \(\ln x = 3\). Give the exact answer.

\(\ln\) is base \(e\), so in exponential form \(x = e^3\). It is positive, so it works. (About 20.09.)

Solve \(\log_2 x + \log_2(x + 2) = 3\).

Condense: \(\log_2\big(x(x + 2)\big) = 3\), so \(x(x + 2) = 2^3 = 8\).

\(x^2 + 2x - 8 = 0\), which factors as \((x + 4)(x - 2) = 0\). So \(x = -4\) or \(x = 2\).

Check: \(x = -4\) gives \(\log_2(-4)\), undefined, so it is extraneous. \(x = 2\): \(\log_2 2 + \log_2 4 = 1 + 2 = 3\). Answer: \(x = 2\).

Solve \(\log_3(x + 6) - \log_3 x = 1\).

Condense with the quotient rule: \(\log_3 \dfrac{x + 6}{x} = 1\), so \(\dfrac{x + 6}{x} = 3^1 = 3\).

Multiply by \(x\): \(x + 6 = 3x\), so \(6 = 2x\) and \(x = 3\).

Check: \(\log_3 9 - \log_3 3 = 2 - 1 = 1\). Answer: \(x = 3\).

A student solves \(\log_4 x + \log_4(x - 6) = 2\) and gets \(x = 8\) and \(x = -2\). Which statement is true?

  1. The only solution is \(x = 8\).
  2. The only solution is \(x = -2\).
  3. Both \(x = 8\) and \(x = -2\) work.
  4. Neither value is a solution.

Check \(x = 8\): the inputs are \(8\) and \(8 - 6 = 2\), both positive. \(\log_4 8 + \log_4 2 = \log_4 16 = 2\). It works.

Check \(x = -2\): \(\log_4(-2)\) is undefined, so \(-2\) is extraneous.

Choice A: the only solution is \(x = 8\).

Step 5 of 5Wrap-up

You can now solve log equations: condense to one log, rewrite in exponential form (or set the insides equal), solve, and cross out any answer that puts a zero or negative number inside a log.

Read or watch next: OpenStax Intermediate Algebra 2e, 10.5 Solve Exponential and Logarithmic Equations (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Logarithmic equations (short videos). Quick rules: Exponentials and logs cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why do log equations sometimes give extraneous answers?" It knows this course and can explain things another way.