Lösung 3.4:1c
Aus Online Mathematik Brückenkurs 1
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Version vom 14:41, 22. 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,
| \displaystyle \ln\bigl(3e^x\bigr) = \ln\bigl(7\cdot 2^x\bigr)\,\textrm{,} |
and use the log laws to make \displaystyle x more accessible,
| \displaystyle \ln 3 + x\cdot \ln e = \ln 7 + x\cdot \ln 2\,\textrm{.} |
Then, collect together the \displaystyle x terms on the left-hand side,
| \displaystyle x(\ln e-\ln 2) = \ln 7-\ln 3\,\textrm{.} |
The solution is now
| \displaystyle x = \frac{\ln 7-\ln 3}{\ln e-\ln 2} = \frac{\ln 7-\ln 3}{1-\ln 2}\,\textrm{.} |
