Why is $-\ \frac{1}{2}\ln(\frac{1}{9})$ equal to $\frac{\ln(9)}{2}$?

Question

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.

Answer 1

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 $-\frac{1}{2}\ln(\frac{1}{9})=\frac{\ln(9)}{2}$

Answer 2

$\ln (1/9) = \ln (9^{-1})=-1 \cdot \ln (9)$

Answer 3

$\ln(\frac{1}{9})=\ln(9^{-1})=(-1)\ln(9)$ ;)

Answer 4

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.