From Förberedande kurs i matematik 1

Revision as of 09:28, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:5e moved to Solution 3.3:5e: Robot: moved page) ← Previous diff		Current revision (07:33, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The argument of ln can be written as
-	<center> [[Image:3_3_5e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\frac{1}{e^{2}} = e^{-2}</math>}}
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		+	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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Current revision

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{.}