Lösung 3.3:5a

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The notation ln is used for the logarithm with the natural base <math>e = 2\textrm{.}7182818\ldots\,,</math> and we therefore have that <math>\ln e = 1\,</math>, which together with the logarithm law, <math>b\lg a=\lg a^b</math>, gives that
The notation ln is used for the logarithm with the natural base <math>e = 2\textrm{.}7182818\ldots\,,</math> and we therefore have that <math>\ln e = 1\,</math>, which together with the logarithm law, <math>b\lg a=\lg a^b</math>, gives that
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{{Displayed math||<math>\ln e^3 + \ln e^2 = 3\cdot\ln e + 2\cdot \ln e = 3\cdot 1 + 2\cdot 1 = 5\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\ln e^3 + \ln e^2 = 3\cdot\ln e + 2\cdot \ln e = 3\cdot 1 + 2\cdot 1 = 5\,\textrm{.}</math>}}

Version vom 08:44, 22. Okt. 2008

The notation ln is used for the logarithm with the natural base \displaystyle e = 2\textrm{.}7182818\ldots\,, and we therefore have that \displaystyle \ln e = 1\,, which together with the logarithm law, \displaystyle b\lg a=\lg a^b, gives that

\displaystyle \ln e^3 + \ln e^2 = 3\cdot\ln e + 2\cdot \ln e = 3\cdot 1 + 2\cdot 1 = 5\,\textrm{.}