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Study Guides > College Algebra

Evaluate Logarithms

Learning Objectives

  • Evaluate logarithms with and without a calculator
  • Evaluate logarithms with base 10, and base e
Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\mathrm{log}}_{2}8[/latex]. We ask, "To what exponent must 2 be raised in order to get 8?" Because we already know [latex]{2}^{3}=8[/latex], it follows that [latex]{\mathrm{log}}_{2}8=3[/latex]. Now consider solving [latex]{\mathrm{log}}_{7}49[/latex] and [latex]{\mathrm{log}}_{3}27[/latex] mentally.
  • We ask, "To what exponent must 7 be raised in order to get 49?" We know [latex]{7}^{2}=49[/latex]. Therefore, [latex]{\mathrm{log}}_{7}49=2[/latex]
  • We ask, "To what exponent must 3 be raised in order to get 27?" We know [latex]{3}^{3}=27[/latex]. Therefore, [latex]{\mathrm{log}}_{3}27=3[/latex]
Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate [latex]{\mathrm{log}}_{\frac{2}{3}}\frac{4}{9}[/latex] mentally.
  • We ask, "To what exponent must [latex]\frac{2}{3}[/latex] be raised in order to get [latex]\frac{4}{9}[/latex]? " We know [latex]{2}^{2}=4[/latex] and [latex]{3}^{2}=9[/latex], so [latex]{\left(\frac{2}{3}\right)}^{2}=\frac{4}{9}[/latex]. Therefore, [latex]{\mathrm{log}}_{\frac{2}{3}}\left(\frac{4}{9}\right)=2[/latex].

How To: Given a logarithm of the form [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex], evaluate it mentally.

  1. Rewrite the argument x as a power of b: [latex]{b}^{y}=x[/latex].
  2. Use previous knowledge of powers of b identify y by asking, "To what exponent should b be raised in order to get x?"

Example: Solving Logarithms Mentally

Solve [latex]y={\mathrm{log}}_{4}\left(64\right)[/latex] without using a calculator.

Answer: First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[/latex]. Next, we ask, "To what exponent must 4 be raised in order to get 64?" We know [latex]{4}^{3}=64[/latex] Therefore, [latex-display]\mathrm{log}{}_{4}\left(64\right)=3[/latex-display]

Try It

Solve [latex]y={\mathrm{log}}_{121}\left(11\right)[/latex] without using a calculator.

Answer: [latex]{\mathrm{log}}_{121}\left(11\right)=\frac{1}{2}[/latex] (recalling that [latex]\sqrt{121}={\left(121\right)}^{\frac{1}{2}}=11[/latex] )

Example: Evaluating the Logarithm of a Reciprocal

Evaluate [latex]y={\mathrm{log}}_{3}\left(\frac{1}{27}\right)[/latex] without using a calculator.

Answer: First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\frac{1}{27}[/latex]. Next, we ask, "To what exponent must 3 be raised in order to get [latex]\frac{1}{27}[/latex]"? We know [latex]{3}^{3}=27[/latex], but what must we do to get the reciprocal, [latex]\frac{1}{27}[/latex]? Recall from working with exponents that [latex]{b}^{-a}=\frac{1}{{b}^{a}}[/latex]. We use this information to write

[latex]\begin{array}{l}{3}^{-3}=\frac{1}{{3}^{3}}\hfill \\ =\frac{1}{27}\hfill \end{array}[/latex]

Therefore, [latex]{\mathrm{log}}_{3}\left(\frac{1}{27}\right)=-3[/latex].

Try It

Evaluate [latex]y={\mathrm{log}}_{2}\left(\frac{1}{32}\right)[/latex] without using a calculator.

Answer: [latex]{\mathrm{log}}_{2}\left(\frac{1}{32}\right)=-5[/latex]

Use common logarithms

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base b, exponent x, and output y. Then we write [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

Example: Converting from Exponential Form to Logarithmic Form

Write the following exponential equations in logarithmic form.
  1. [latex]{2}^{3}=8[/latex]
  2. [latex]{5}^{2}=25[/latex]
  3. [latex]{10}^{-4}=\frac{1}{10,000}[/latex]

Answer: First, identify the values of b, y, and x. Then, write the equation in the form [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

  1. [latex]{2}^{3}=8[/latex]Here, = 2, = 3, and = 8. Therefore, the equation [latex]{2}^{3}=8[/latex] is equivalent to [latex]{\mathrm{log}}_{2}\left(8\right)=3[/latex].
  2. [latex]{5}^{2}=25[/latex]Here, = 5, = 2, and = 25. Therefore, the equation [latex]{5}^{2}=25[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex].
  3. [latex]{10}^{-4}=\frac{1}{10,000}[/latex]Here, = 10, = –4, and [latex]y=\frac{1}{10,000}[/latex]. Therefore, the equation [latex]{10}^{-4}=\frac{1}{10,000}[/latex] is equivalent to [latex]{\text{log}}_{10}\left(\frac{1}{10,000}\right)=-4[/latex].

Try It

Write the following exponential equations in logarithmic form.
  1. [latex]{3}^{2}=9[/latex]
  2. [latex]{5}^{3}=125[/latex]
  3. [latex]{2}^{-1}=\frac{1}{2}[/latex]

Answer:

  1. [latex]{3}^{2}=9[/latex] is equivalent to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
  2. [latex]{5}^{3}=125[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
  3. [latex]{2}^{-1}=\frac{1}{2}[/latex] is equivalent to [latex]{\text{log}}_{2}\left(\frac{1}{2}\right)=-1[/latex]

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  • Question ID 35022. Authored by: Smart,Jim, mb Sousa,James. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].