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.

There exists the following property for logarithms:
$$n \ln{x} = \ln{x^n}$$
So for your problem you have:
$$ -\frac{1}{2} \ln{\left(\frac{1}{9}\right)}=\frac{1}{2}\ln{\left(\left(\frac{1}{9}\right)^{-1}\right)}=\frac{1}{2}\ln{9}= \frac{\ln9}{2}$$
I hope this is sufficient as an explanation.
