Lösung 3.3:3a

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By writing the argument
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By writing the argument <math>8</math> as <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^3</math>, the logarithm law, <math>\lg a^b = b\lg a</math>, gives
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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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{{Displayed math||<math>\log _{2}8 = \log _{2} 2^3 = 3\cdot\log _{2}2 = 3\cdot 1 = 3\,,</math>}}
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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 <math>\log _{2}2 = 1\,</math>.
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where we have used
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<math>\log _{2}2=1</math>.
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Version vom 06:24, 2. Okt. 2008

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

Vorlage:Displayed math

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“