Lösung 3.3:5b

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By using the logarithm laws,
By using the logarithm laws,
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{{Displayed math||<math>\begin{align}
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<math>\lg a+\lg b=\lg \left( a\centerdot b \right)</math>
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\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt]
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\ln a - \ln b &= \ln\frac{a}{b}\,,
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\end{align}</math>}}
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<math>\text{log }a-\text{ log }b=\text{log}\left( \frac{a}{b} \right)</math>
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we can collect together the terms into one logarithmic expression
we can collect together the terms into one logarithmic expression
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{{Displayed math||<math>\begin{align}
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\ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt]
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&= \ln 8 - \ln(4\cdot 2)\\[5pt]
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&= \ln\frac{8}{4\cdot 2}\\[5pt]
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&= \ln 1\\[5pt]
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&= 0\,,
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\end{align}</math>}}
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<math>\begin{align}
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where <math>\ln 1 = 0</math>, since <math>e^{0}=1</math> (the equality <math>a^{0}=1</math> holds for all <math>a\ne 0</math>).
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& \ln 8-\ln 4-\ln 2=\ln 8-\left( \ln 4+\ln 2 \right)=\ln 8-\ln \left( 4\centerdot 2 \right) \\
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& =\ln \frac{8}{4\centerdot 2}=\ln 1=0, \\
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\end{align}</math>
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where ln 1 =0, since
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<math>e^{0}=1</math>
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(the equality
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<math>a^{0}=1</math>
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holds for all
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<math>a\ne 0</math>
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).
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Version vom 07:27, 2. Okt. 2008

By using the logarithm laws,

Vorlage:Displayed math

we can collect together the terms into one logarithmic expression

Vorlage:Displayed math

where \displaystyle \ln 1 = 0, since \displaystyle e^{0}=1 (the equality \displaystyle a^{0}=1 holds for all \displaystyle a\ne 0).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:5b“