Lösung 3.3:3g

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Using the logarithm law, <math>\lg a-\lg b = \lg\frac{a}{b}\,</math>, the expression can be calculated as
Using the logarithm law, <math>\lg a-\lg b = \lg\frac{a}{b}\,</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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{{Abgesetzte Formel||<math>\log_3 12 - \log_3 4 = \log_3\frac{12}{4} = \log _3 3 = 1\,\textrm{.}</math>}}
Another way is to write <math>12 = 3\cdot 4</math> and use the logarithm law,
Another way is to write <math>12 = 3\cdot 4</math> and use the logarithm law,
<math>\lg (ab) = \lg a + \lg b\,</math>,
<math>\lg (ab) = \lg a + \lg b\,</math>,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\log _{3}12 - \log _{3}4
\log _{3}12 - \log _{3}4
&= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt]
&= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt]

Version vom 08:43, 22. Okt. 2008

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

\displaystyle \log_3 12 - \log_3 4 = \log_3\frac{12}{4} = \log _3 3 = 1\,\textrm{.}

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

\displaystyle \begin{align}

\log _{3}12 - \log _{3}4 &= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt] &= \log_{3}3 + \log _{3}4 - \log _{3}4\\[5pt] &= \log _{3}3\\[5pt] &= 1\,\textrm{.} \end{align}

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