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

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

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

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.