Lösung 3.3:5e

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The argument of ln can be written as
The argument of ln can be written as
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{{Displayed math||<math>\frac{1}{e^{2}} = e^{-2}</math>}}
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{{Abgesetzte Formel||<math>\frac{1}{e^{2}} = e^{-2}</math>}}
and with the logarithm law, <math>\ln a^{b} = b\ln a</math>, we obtain
and with the logarithm law, <math>\ln a^{b} = b\ln a</math>, we obtain
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{{Displayed math||<math>\ln \frac{1}{e^{2}} = \ln e^{-2} = (-2)\cdot\ln e = (-2)\cdot 1 = -2\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\ln \frac{1}{e^{2}} = \ln e^{-2} = (-2)\cdot\ln e = (-2)\cdot 1 = -2\,\textrm{.}</math>}}

Version vom 08:44, 22. Okt. 2008

The argument of ln can be written as

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

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

\displaystyle \ln \frac{1}{e^{2}} = \ln e^{-2} = (-2)\cdot\ln e = (-2)\cdot 1 = -2\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:5e“