Lösung 3.4:1c

Aus Online Mathematik Brückenkurs 1

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The equation has the same form as the equation in exercise c and we can therefore use the same strategy.
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The equation has the same form as the equation in exercise b and we can therefore use the same strategy.
First, we take logs of both sides,
First, we take logs of both sides,
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{{Displayed math||<math>\ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,}</math>}}
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<math>\ln \left( 3e^{x} \right)=\ln \left( 7\centerdot 2^{x} \right)</math>
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and use the log laws to make <math>x</math> more accessible,
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{{Displayed math||<math>\ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.}</math>}}
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and use the log laws to make
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Then, collect together the <math>x</math> terms on the left-hand side,
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<math>x</math>
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more accessible:
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<math>\ln 3+x\centerdot \ln e=\ln 7+x\centerdot \ln 2</math>
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Then, collect together the <math>x</math> terms on the left-hand side:
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<math>x\left( \ln e-\ln 2 \right)=\ln 7-\ln 3</math>
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{{Displayed math||<math>x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.}</math>}}
The solution is now
The solution is now
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{{Displayed math||<math>x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.}</math>}}
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<math>x=\frac{\ln 7-\ln 3}{\ln e-\ln 2}=\frac{\ln 7-\ln 3}{1-\ln 2}</math>
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Version vom 10:50, 2. Okt. 2008

The equation has the same form as the equation in exercise b and we can therefore use the same strategy.

First, we take logs of both sides,

Vorlage:Displayed math

and use the log laws to make \displaystyle x more accessible,

Vorlage:Displayed math

Then, collect together the \displaystyle x terms on the left-hand side,

Vorlage:Displayed math

The solution is now

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.4:1c“