Showing $ (2/3)^{\log_{3/2}(n/n_0)} =n_0/n $

Question

I read the following identity in the CLRS Introduction to Algorithms book, and I can't work out the computation. $$ (2/3)^{\log_{3/2}(n/n_0)} =n_0/n $$ I tried to expand the exponent using the division rule for logarithms and bring the factor with the negative power down as a denominator. I couldn't figure out what I could do next.

Answer 1

Applying the hint given by Surb leads to

$$\left(\frac{1}{3/2}\right)^{\log_{3/2}(n/n_0)} = \left(\frac{1^{\log_{3/2}(n/n_0)}}{\left(\dfrac{3}{2}\right)^{\log_{3/2}(n/n_0)}}\right) = \frac{1}{\left(\dfrac{3}{2}\right)^{\log_{3/2}(n/n_0)}} = \frac{1}{\dfrac{n}{n_0}} = \frac{n_0}{n}, $$ as desired.

Thanks Surb.