Lösung 3.3:3g

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Using the logarithm law,
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<center> [[Image:3_3_3g.gif]] </center>
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<math>\lg a-\lg b=\lg \left( \frac{a}{b} \right)</math>, the expression can be calculated as
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<math>\log _{3}12-\log _{3}4=\log _{3}\frac{12}{4}=\log _{3}3=1</math>
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Another way is to write
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<math>\text{12}=\text{3}\centerdot \text{4 }</math>
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and use the logarithm law,
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<math>\lg \left( ab \right)=\lg a+\lg b</math>,
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<math>\begin{align}
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& \log _{3}12-\log _{3}4=\log _{3}\left( 3\centerdot 4 \right)-\log _{3}4 \\
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& =\log _{3}3+\log _{3}4-\log _{3}4=\log _{3}3=1 \\
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\end{align}</math>

Version vom 14:33, 25. Sep. 2008

Using the logarithm law, \displaystyle \lg a-\lg b=\lg \left( \frac{a}{b} \right), the expression can be calculated as


\displaystyle \log _{3}12-\log _{3}4=\log _{3}\frac{12}{4}=\log _{3}3=1


Another way is to write \displaystyle \text{12}=\text{3}\centerdot \text{4 } and use the logarithm law, \displaystyle \lg \left( ab \right)=\lg a+\lg b,


\displaystyle \begin{align} & \log _{3}12-\log _{3}4=\log _{3}\left( 3\centerdot 4 \right)-\log _{3}4 \\ & =\log _{3}3+\log _{3}4-\log _{3}4=\log _{3}3=1 \\ \end{align}

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