Lösung 3.3:2g

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We know that
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<center> [[Image:3_3_2g.gif]] </center>
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<math>10^{\lg x}=x</math>, so therefore we rewrite the exponent as
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<math>-\lg 0.1=\left( -1 \right)\centerdot \lg 0.1=\lg 0.1^{-1}</math>
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by using the log law
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<math>b\lg a=\lg a^{b}</math>. This gives
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<math>10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10</math>

Version vom 13:31, 25. Sep. 2008

We know that \displaystyle 10^{\lg x}=x, so therefore we rewrite the exponent as \displaystyle -\lg 0.1=\left( -1 \right)\centerdot \lg 0.1=\lg 0.1^{-1} by using the log law \displaystyle b\lg a=\lg a^{b}. This gives


\displaystyle 10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10

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