Lösung 3.3:3b

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Because we are working with <math>\log _{9}</math>, we express 1/3 as a power of 9,
Because we are working with <math>\log _{9}</math>, we express 1/3 as a power of 9,
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{{Displayed math||<math>\frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<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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{{Abgesetzte Formel||<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>}}

Version vom 08:42, 22. Okt. 2008

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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:3b“