I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
Then ln(8)=-x
So x=-ln(8)
Apparently the answer is then x=ln(1/8)
I just don't know why the negative ln8 is equal to 1/8.
That last line is not true. You need to use the following facts:
$\displaystyle \ln{1} = 0$
$\displaystyle \ln{a} - \ln{b} = \ln{\left(\frac{a}{b}\right)}$
Then
$\displaystyle - \ln{8} = 0 - \ln{8} = \ln{1} - \ln{8} = \ln{\left(\frac{1}{8}\right)}$
edit: oops, beaten to it! My Latex skills are too slow.
The technical reason...is well technical. But think about it like this, $\displaystyle \ln(x)$ is just the solution to $\displaystyle e^y=x$. So let $\displaystyle \alpha=\ln\left(\frac{1}{8}\right)$ then $\displaystyle e^{\alpha}=\frac{1}{8}$ or equivalently $\displaystyle 8=\frac{1}{e^{\alpha}}=e^{-\alpha}$ so that $\displaystyle \ln(8)=-\alpha\implies -\ln(8)=\alpha=\ln\left(\tfrac{1}{8}\right)$
Of course neither $\displaystyle -ln(8)$ nor $\displaystyle ln\left(\frac{1}{8}\right)$ are fully simplified
$\displaystyle -3ln(2) = 3ln\left(\frac{1}{2}\right)$