How do I take a fraction to a negative power? I ran into this issue during my homework. Using the rules of logarithms, I need to prove that 
$$
-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln3-\ln2
$$
So here were my steps:


*

*First step:
$$
-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln\left(\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}\right)
$$
And that's as far as I got, because now I want to use the form $\ln(a/b) = \ln(a) - \ln(b)$, but first I need to reduce the fraction because it is raised to the $-2$.


How do I evaluate $\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}$ ?
Thanks

By definition $$a^{-k} = \frac 1{a^k}$$
So  $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$
$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$
It will help to realize that $(\frac ab)^{-1} = 1/(a/b) = \frac ba$ and that $(\frac ab)^k = \frac {a^k}{b^k}$ to realize that that means $$\left(\frac ab\right)^{-k} = \frac 1{\left(\frac ab\right)^k}= \frac 1{\left(\frac {a^k}{b^k}\right)} = \frac {b^k}{a^k}.$$
(Also $(\frac ab)^{-k} = [(\frac ab)^{-1}]^k = (\frac ba)^k=\frac {b^k}{a^k}$ or that $(\frac ab)^{-k} = \frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \frac {b^k}{a^k}$.)
In any event
$$\left(\frac {2}{\sqrt 6}\right)^{-2} = \left(\frac {\sqrt 6}{ 2}\right)^2 = \frac {\sqrt 6^2}{2^2} = \frac 64 = \frac 32.$$
