Lösung 3.4:1a
Aus Online Mathematik Brückenkurs 1
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Version vom 14:40, 22. Okt. 2008
We solve an equation of this type by taking the natural logarithm of both sides
| \displaystyle \ln e^x = \ln 13\,\textrm{.} |
Then, using the log law \displaystyle \ln a^{b} = b\cdot \ln a, the unknown \displaystyle x can then be moved down as a factor on the left-hand side
| \displaystyle x\cdot \ln e = \ln 13 |
and then it is simple to solve for \displaystyle x,
| \displaystyle x = \frac{\ln 13}{\ln e} = \frac{\ln 13}{1} = \ln 13\,\textrm{.} |
Note: One thing that we haven't mentioned in this solution is that before we carry out the first step and take the logarithm of both sides of the equation, it is necessary to be certain that the left- and right-hand sides are positive (it is not possible to take the logarithm of a negative number). Because the equation is
| \displaystyle e^x=13 |
we see directly that the right-hand side is positive. The left-hand side is also positive because “a positive number \displaystyle e\approx 2\textrm{.}718 raised to a number” always gives a positive number.
