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Lösung 3.3:3c

Aus Online Mathematik Brückenkurs 1

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K (Lösning 3.3:3c moved to Solution 3.3:3c: Robot: moved page)
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First, we rewrite the number
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<center> [[Image:3_3_3c.gif]] </center>
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<math>0.\text{125 }</math>
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as a fraction which we also simplify:
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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
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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>

Version vom 14:13, 25. Sep. 2008

First, we rewrite the number 0125 as a fraction which we also simplify:


0125 =125 1000=103525=555253=123=2−3


Because 0125 was expressed as a power of 2, the logarithm can be calculated in full:


log20125 =log22−3=−3log22=−31=−3 

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