Lösung 3.3:3d

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We write the argument of
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<center> [[Image:3_3_3d.gif]] </center>
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<math>\log _{3}</math>
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as a power of
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<math>\text{3}</math>,
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<math>9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}</math>
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and then simplify the expression with the logarithm laws:
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<math>\log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.</math>

Version vom 14:19, 25. Sep. 2008

We write the argument of \displaystyle \log _{3} as a power of \displaystyle \text{3},


\displaystyle 9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}


and then simplify the expression with the logarithm laws:


\displaystyle \log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.

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