Lösung 3.3:3g

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Using the logarithm law,
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Using the logarithm law, <math>\lg a-\lg b = \lg\frac{a}{b}\,</math>, the expression can be calculated as
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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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{{Displayed math||<math>\log_3 12 - \log_3 4 = \log_3\frac{12}{4} = \log _3 3 = 1\,\textrm{.}</math>}}
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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 <math>12 = 3\cdot 4</math> and use the logarithm law,
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<math>\lg (ab) = \lg a + \lg b\,</math>,
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{{Displayed math||<math>\begin{align}
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Another way is to write
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\log _{3}12 - \log _{3}4
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<math>\text{12}=\text{3}\centerdot \text{4 }</math>
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&= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt]
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and use the logarithm law,
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&= \log_{3}3 + \log _{3}4 - \log _{3}4\\[5pt]
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<math>\lg \left( ab \right)=\lg a+\lg b</math>,
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&= \log _{3}3\\[5pt]
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&= 1\,\textrm{.}
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\end{align}</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>
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Version vom 07:00, 2. Okt. 2008

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

Vorlage:Displayed math

Another way is to write \displaystyle 12 = 3\cdot 4 and use the logarithm law, \displaystyle \lg (ab) = \lg a + \lg b\,,

Vorlage:Displayed math

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