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Solution 3.4:1c

From Förberedande kurs i matematik 1

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m (Lösning 3.4:1c moved to Solution 3.4:1c: Robot: moved page)
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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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<center> [[Image:3_4_1c.gif]] </center>
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First, we take logs of both sides,
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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
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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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The solution is now
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<math>x=\frac{\ln 7-\ln 3}{\ln e-\ln 2}=\frac{\ln 7-\ln 3}{1-\ln 2}</math>

Revision as of 13:22, 12 September 2008

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

First, we take logs of both sides,


ln3ex=ln72x 


and use the log laws to make x more accessible:


ln3+xlne=ln7+xln2


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


xlneln2=ln7ln3 


The solution is now


x=lneln2ln7ln3=1ln2ln7ln3