Solution 3.3:5e

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.3:5e moved to Solution 3.3:5e: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
The argument of ln can be written as
-
<center> [[Image:3_3_5e.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\frac{1}{e^{2}}=e^{-2}</math>
 +
 
 +
 
 +
and with the logarithm law,
 +
<math>\lg a^{b}=b\lg a</math>, we obtain
 +
 
 +
 
 +
<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


\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

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.3:5e"