Alternative solution to this? $$\displaystyle a^{-\frac{2}{3}} = 9$$
I've solved it by multiplying the reciprocal but how do you solve it by taking $\frac{1}{a}$ to the power of $\frac{2}{3} = 9$ and so forth?

$$a^{-2/3}=9=3^2\implies a^{-2}=(3^2)^3=3^6\iff a^2=3^{-6}$$
Extract square root in both sides
