Lösung 3.3:3a

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By writing the argument
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<math>\text{8}</math>
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as
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<math>8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3}</math>, the logarithm law,
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<math>\lg a^{b}=b\lg a</math>, gives
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<math>\log _{2}8=\log _{2}2^{3}=3\centerdot \log _{2}2=3\centerdot 1=3</math>
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where we have used
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<math>\log _{2}2=1</math>.

Version vom 13:57, 25. Sep. 2008

By writing the argument \displaystyle \text{8} as \displaystyle 8=2\centerdot 4=2\centerdot 2\centerdot 2=2^{3}, the logarithm law, \displaystyle \lg a^{b}=b\lg a, gives


\displaystyle \log _{2}8=\log _{2}2^{3}=3\centerdot \log _{2}2=3\centerdot 1=3


where we have used \displaystyle \log _{2}2=1.

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