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

Well you may start by distributing the index since $2$ and $\sqrt 6$ are positive. Thus $$\left(\frac{2}{\sqrt{6}}\right)^{-2}=\frac{2^{-2}}{{\sqrt 6}^{-2}}.$$
Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=\frac {1}{a^n}.$$ Applying this to your expression, we have $$\frac{2^{-2}}{{\sqrt 6}^{-2}}=\frac{\frac {1}{2^2}}{\frac{1}{{\sqrt 6}^2}}=\frac{\frac {1}{4}}{\frac{1}{6}}=\frac{6}{4}=\frac 32.$$
