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Solution 3.3:5b

From Förberedande kurs i matematik 1

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m (Lösning 3.3:5b moved to Solution 3.3:5b: Robot: moved page)
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By using the logarithm laws,
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<center> [[Image:3_3_5b.gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\lg a+\lg b=\lg \left( a\centerdot b \right)</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
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<math>\begin{align}
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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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).

Revision as of 08:26, 26 September 2008

By using the logarithm laws,


lga+lgb=lgab 


log a log b=logba 

we can collect together the terms into one logarithmic expression


ln8ln4ln2=ln8ln4+ln2=ln8ln42=ln842=ln1=0


where ln 1 =0, since e0=1 (the equality a0=1 holds for all a=0 ).