Lösung 3.3:3b

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Because we are working with
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<center> [[Image:3_3_3b.gif]] </center>
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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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Using the logarithm laws, we get
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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>

Version vom 14:06, 25. Sep. 2008

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


\displaystyle \frac{1}{3}=\frac{1}{\sqrt{9}}=\frac{1}{9^{{1}/{2}\;}}=9^{-{1}/{2}\;}


Using the logarithm laws, we get


\displaystyle \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}.

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