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Step 13: Algebra 2

Exponentials and logarithms

Step 13 · Algebra 2 topics

Read logarithms as exponents, switch between exponential and log form, use the log rules, and solve exponential equations with matching bases or with logs.

Logarithms undo exponents

A logarithm answers the question "what power?". log⁡28=3\log_{2} 8 = 3 because 2³ = 8. In general, log⁡bx=y\log_{b} x = y means by=xb^{y} = x.

Exponential formLogarithmic form
2³ = 8log⁡28=3\log_{2} 8 = 3
10² = 100log⁡10100=2\log_{10} 100 = 2
5⁰ = 1log⁡51=0\log_{5} 1 = 0
91/2=39^{1/2} = 3log⁡93=12\log_{9} 3 = \frac{1}{2}

Example: Evaluating

Find log⁡381\log_{3} 81 and log⁡218\log_{2} \frac{1}{8}.

  1. 3⁴ = 81, so the first is 4.
  2. 2⁻³ is 18\frac{1}{8}, so the second is −3.

Answer4 and −3

y = 2ˣ and y = log₂ x are mirror images in the dashed line y = x: a log undoes an exponent.

Log rules

Because logs are exponents, the exponent rules become log rules.

RuleExample
log⁡b(MN)=log⁡bM+log⁡bN\log_{b}(MN) = \log_{b} M + \log_{b} Nlog⁡232=log⁡24+log⁡28\log_{2} 32 = \log_{2} 4 + \log_{2} 8
log⁡bMN=log⁡bM−log⁡bN\log_{b}\frac{M}{N} = \log_{b} M - \log_{b} Nlog⁡39=log⁡327−log⁡33\log_{3} 9 = \log_{3} 27 - \log_{3} 3
log⁡b(Mk)=klog⁡bM\log_{b}(M^{k}) = k\log_{b} Mlog⁡1000=3log⁡10=3\log 1000 = 3\log 10 = 3

Example: Expanding

Expand log⁡5x3y\log_{5}\frac{x^{3}}{y}.

  1. Use the quotient rule, then the power rule.

Answer3log⁡5x−log⁡5y3\log_{5} x - \log_{5} y

Example: Condensing

Write 2log⁡x+log⁡52\log x + \log 5 as a single log.

  1. The power rule turns 2 log x into log x².
  2. The product rule joins the two logs.

Answerlog⁡(5x2)\log(5x^{2})

Solving exponential equations

To solve an exponential equation, write both sides with the same base if you can. If not, take the log of both sides.

Example: The same base

Solve 4ˣ = 32.

  1. 4 = 2² and 32 = 2⁵, so 2²ˣ = 2⁵.
  2. 2x = 5.

Answerx = 2.5

Example: Using logs

Solve 3ˣ = 20, to two decimal places.

  1. Take logs of both sides: x log 3 = log 20.
  2. x=log⁡20log⁡3x = \frac{\log 20}{\log 3}, about 1.301 ÷ 0.477.

Answerx ≈ 2.73

Example: Doubling time

$500 grows at 6% a year. When does it reach $1,000?

  1. 500 × 1.06ᵗ = 1,000, so 1.06ᵗ = 2.
  2. t=log⁡2log⁡1.06t = \frac{\log 2}{\log 1.06}, which is about 11.9.

AnswerAfter about 11.9 years

Try it yourself

Work each one on paper first, then open the answer to check.

  1. What is log⁡464\log_{4} 64?

    Show answer

    3

  2. What is log⁡100.01\log_{10} 0.01?

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    −2

  3. Write 7² = 49 in log form.

    Show answer

    log⁡749=2\log_{7} 49 = 2

  4. Expand log⁡(10x)\log(10x).

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    1+log⁡x1 + \log x

  5. Solve log⁡2x=5\log_{2} x = 5.

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    x = 32

  6. Solve 9ˣ = 27.

    Show answer

    x = 1.5

  7. Solve 2ˣ = 10, to two decimal places.

    Show answer

    x ≈ 3.32

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