Lösung 3.4:1b
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Version vom 14:40, 22. Okt. 2008
In the equation, both sides are positive because the factors \displaystyle e^{x} and \displaystyle 3^{-x} are positive regardless of the value of \displaystyle x (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,
| \displaystyle \ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.} |
Using the log laws, we can divide up the products into several logarithmic terms,
| \displaystyle \ln 13+\ln e^{x} =\ln 2+\ln 3^{-x}, |
and using the law \displaystyle \ln a^{b}=b\cdot \ln a, we can get rid of \displaystyle x from the exponents
| \displaystyle \ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.} |
Collect \displaystyle x on one side and the other terms on the other,
| \displaystyle x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.} |
Take out \displaystyle x on the left-hand side and use \displaystyle \ln e=1,
| \displaystyle x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.} |
Then, solve for \displaystyle x,
| \displaystyle x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.} |
Note: Because \displaystyle \ln 2 < \ln 13, we can write the answer as
| \displaystyle x=-\frac{\ln 13-\ln 2}{1+\ln 3} |
to indicate that \displaystyle x is negative.
