Lösung 3.3:4c

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All three arguments of the logarithm can be written as powers of
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<center> [[Image:3_3_4c.gif]] </center>
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<math>\text{3}</math>,
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<math>\begin{align}
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& 27^{\frac{1}{3}}=\left( 3^{3} \right)^{\frac{1}{3}}=3^{3\centerdot \frac{1}{3}}=3^{1}=3, \\
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& \frac{1}{9}=\frac{1}{3^{2}}=3^{-2} \\
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\end{align}</math>
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and it is therefore appropriate to use base
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<math>\text{3}</math>
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when simplifying using the logarithms, even if we have the base
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<math>\text{1}0</math>
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logarithm, lg,
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<math>\begin{align}
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& \lg 27^{\frac{1}{3}}+\frac{\lg 3}{2}+\lg \frac{1}{9}=\lg 3+\frac{1}{2}\lg 3+\lg 3^{-2} \\
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& =\lg 3+\frac{1}{2}\lg 3+\left( -2 \right)\centerdot \lg 3 \\
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& =\left( 1+\frac{1}{2}-2 \right)\lg 3=-\frac{1}{2}\lg 3 \\
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\end{align}</math>
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This expression cannot be simplified any further.

Version vom 14:58, 25. Sep. 2008

All three arguments of the logarithm can be written as powers of \displaystyle \text{3},

\displaystyle \begin{align} & 27^{\frac{1}{3}}=\left( 3^{3} \right)^{\frac{1}{3}}=3^{3\centerdot \frac{1}{3}}=3^{1}=3, \\ & \frac{1}{9}=\frac{1}{3^{2}}=3^{-2} \\ \end{align}


and it is therefore appropriate to use base \displaystyle \text{3} when simplifying using the logarithms, even if we have the base \displaystyle \text{1}0 logarithm, lg,


\displaystyle \begin{align} & \lg 27^{\frac{1}{3}}+\frac{\lg 3}{2}+\lg \frac{1}{9}=\lg 3+\frac{1}{2}\lg 3+\lg 3^{-2} \\ & =\lg 3+\frac{1}{2}\lg 3+\left( -2 \right)\centerdot \lg 3 \\ & =\left( 1+\frac{1}{2}-2 \right)\lg 3=-\frac{1}{2}\lg 3 \\ \end{align}


This expression cannot be simplified any further.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:4c“