Why is $-\ \frac{1}{2}\ln(\frac{1}{9})$ equal to $\frac{\ln(9)}{2}$? I solved this problem in my textbook but noticed their solution was different than mine. 
$1. \ 9e^{-2x}=1$ 
$2. \ e^{-2x}=\frac{1}{9}$
$3. -2x=\ln(\frac{1}{9})$
$4. \ x=-\ \frac{1}{2}\ln(\frac{1}{9})$
However, the answer that my textbook gives is $\frac{\ln(9)}{2}$ 
I plugged these expressions into my calculator and they are indeed equivalent, however I don't see what properties I could use to get from my messy answer to the textbook's much cleaner one. Any help would be greatly appreciated. Thank you.

Note that $$\ln x +\ln y =\ln xy, \; \ln 1=\ln e^{0}=0$$
If $x, y$ are positive reals, as seen here. From this, $$\ln x +\ln \frac{1}{x}=0 \iff \ln x =-\ln \frac{1}{x}$$
So $$\ln \frac{1}{9}=-\ln 9$$
So $-\frac{1}{2}\ln(\frac{1}{9})=\frac{\ln(9)}{2}$
