Lösung 3.3:5e

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The argument of ln can be written as
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<center> [[Image:3_3_5e.gif]] </center>
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<math>\frac{1}{e^{2}}=e^{-2}</math>
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and with the logarithm law,
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<math>\lg a^{b}=b\lg a</math>, we obtain
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<math>\ln \frac{1}{e^{2}}=\ln e^{-2}=\left( -2 \right)\centerdot \ln e=\left( -2 \right)\centerdot 1=-2</math>

Version vom 08:45, 26. Sep. 2008

The argument of ln can be written as


\displaystyle \frac{1}{e^{2}}=e^{-2}


and with the logarithm law, \displaystyle \lg a^{b}=b\lg a, we obtain


\displaystyle \ln \frac{1}{e^{2}}=\ln e^{-2}=\left( -2 \right)\centerdot \ln e=\left( -2 \right)\centerdot 1=-2

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