Lösung 3.4:3a
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Version vom 14:42, 22. Okt. 2008
Both left- and right-hand sides are positive for all values of x and this means that we can take the logarithm of both sides and get a more manageable equation,
| \displaystyle \begin{align}
\text{LHS} &= \ln 2^{-x^{2}} = -x^{2}\cdot \ln 2\,,\\[5pt] \text{RHS} &= \ln \bigl(2e^{2x}\bigr) = \ln 2 + \ln e^{2x} = \ln 2 + 2x\cdot \ln e = \ln 2 + 2x\cdot 1\,\textrm{.} \end{align} |
After a little rearranging, the equation becomes
| \displaystyle x^{2}+\frac{2}{\ln 2}x+1=0\,\textrm{.} |
We complete the square of the left-hand side,
| \displaystyle \Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} - \Bigl(\frac{1}{\ln 2} \Bigr)^{2} + 1 = 0\,, |
and move the constant terms over to the right-hand side,
| \displaystyle \Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} = \Bigl(\frac{1}{\ln 2} \Bigr)^{2} - 1\,\textrm{.} |
It can be difficult to see whether the right-hand side is positive or not, but if we remember that \displaystyle e > 2 and that thus \displaystyle \ln 2 < \ln e = 1\,, we must have that \displaystyle (1/\ln 2)^{2} > 1\,, i.e. the right-hand side is positive.
The equation therefore has the solutions
| \displaystyle x=-\frac{1}{\ln 2}\pm \sqrt{\Bigl(\frac{1}{\ln 2} \Bigr)^{2}-1}\,, |
which can also be written as
| \displaystyle x=\frac{-1\pm \sqrt{1-(\ln 2)^{2}}}{\ln 2}\,\textrm{.} |
