Lösung 3.3:2h

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The argument in the logarithm can be rewritten as
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The argument in the logarithm can be rewritten as <math>\frac{1}{10^{2}} = 10^{-2}</math> and then the log law <math>\lg a^b = b\lg a</math> gives the rest
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<math>\frac{1}{10^{2}}=10^{-2}</math>
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and then the log law
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<math>\lg a^{b}=b\lg a</math>
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gives the rest:
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{{Displayed math||<math>\lg \frac{1}{10^2} = \lg 10^{-2} = (-2)\cdot \lg 10 = (-2)\cdot 1 = -2\,\textrm{.}</math>}}
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<math>\lg \frac{1}{10^{2}}=\lg 10^{-2}=\left( -2 \right)\centerdot \lg 10=\left( -2 \right)\centerdot 1=-2</math>
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Version vom 06:22, 2. Okt. 2008

The argument in the logarithm can be rewritten as \displaystyle \frac{1}{10^{2}} = 10^{-2} and then the log law \displaystyle \lg a^b = b\lg a gives the rest

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:2h“