Solution 3.3:3b

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Current revision (06:27, 2 October 2008) (edit) (undo)
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Because we are working with
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Because we are working with <math>\log _{9}</math>, we express 1/3 as a power of 9,
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<math>\log _{9}</math>, we express
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<math>{1}/{3}\;</math>
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as a power of
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<math>\text{9}</math>,
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<math>\frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}</math>
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{{Displayed math||<math>\frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}</math>}}
Using the logarithm laws, we get
Using the logarithm laws, we get
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{{Displayed math||<math>\log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}</math>}}
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<math>\log _{9}\frac{1}{3}=\log _{9}9^{-{1}/{2}\;}=-\frac{1}{2}\centerdot \log _{9}9=-\frac{1}{2}\centerdot 1=-\frac{1}{2}.</math>
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Current revision

Because we are working with \displaystyle \log _{9}, we express 1/3 as a power of 9,

\displaystyle \frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}

Using the logarithm laws, we get

\displaystyle \log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}
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