Lösung 3.3:5b

Aus Online Mathematik Brückenkurs 1

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By using the logarithm laws,
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<center> [[Image:3_3_5b.gif]] </center>
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<math>\lg a+\lg b=\lg \left( a\centerdot b \right)</math>
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<math>\text{log }a-\text{ log }b=\text{log}\left( \frac{a}{b} \right)</math>
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we can collect together the terms into one logarithmic expression
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<math>\begin{align}
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& \ln 8-\ln 4-\ln 2=\ln 8-\left( \ln 4+\ln 2 \right)=\ln 8-\ln \left( 4\centerdot 2 \right) \\
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& =\ln \frac{8}{4\centerdot 2}=\ln 1=0, \\
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\end{align}</math>
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where ln 1 =0, since
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<math>e^{0}=1</math>
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(the equality
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<math>a^{0}=1</math>
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holds for all
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<math>a\ne 0</math>
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).

Version vom 08:26, 26. Sep. 2008

By using the logarithm laws,


\displaystyle \lg a+\lg b=\lg \left( a\centerdot b \right)


\displaystyle \text{log }a-\text{ log }b=\text{log}\left( \frac{a}{b} \right)

we can collect together the terms into one logarithmic expression


\displaystyle \begin{align} & \ln 8-\ln 4-\ln 2=\ln 8-\left( \ln 4+\ln 2 \right)=\ln 8-\ln \left( 4\centerdot 2 \right) \\ & =\ln \frac{8}{4\centerdot 2}=\ln 1=0, \\ \end{align}


where ln 1 =0, since \displaystyle e^{0}=1 (the equality \displaystyle a^{0}=1 holds for all \displaystyle a\ne 0 ).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:5b“