Lösung 3.3:3c

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First, we rewrite the number
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First, we rewrite the number 0.125 as a fraction which we also simplify
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<math>0.\text{125 }</math>
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as a fraction which we also simplify:
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{{Displayed math||<math>0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.}</math>}}
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<math>0.\text{125 }=\frac{\text{125 }}{1000}=\frac{5\centerdot 25}{10^{3}}=\frac{5\centerdot 5\centerdot 5}{\left( 2\centerdot 5 \right)^{3}}=\frac{1}{2^{3}}=2^{-3}</math>
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Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full
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{{Displayed math||<math>\log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.}</math>}}
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Because
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<math>0.\text{125 }</math>
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was expressed as a power of
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<math>\text{2}</math>, the logarithm can be calculated in full:
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<math>\log _{2}0.\text{125 }=\log _{2}2^{-3}=\left( -3 \right)\centerdot \log _{2}2=\left( -3 \right)\centerdot 1=-3</math>
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Version vom 06:33, 2. Okt. 2008

First, we rewrite the number 0.125 as a fraction which we also simplify

Vorlage:Displayed math

Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full

Vorlage:Displayed math

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