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Solution 3.3:5e

From Förberedande kurs i matematik 1

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m (Lösning 3.3:5e moved to Solution 3.3:5e: Robot: moved page)
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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>

Revision as of 08:45, 26 September 2008

The argument of ln can be written as


1e2=e−2


and with the logarithm law, lgab=blga, we obtain


ln1e2=lne−2=−2lne=−21=−2 

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