Logarithms never go amiss

By Vieta Formula, we get: $p+q = 6$ and $pq = 24$ . Now we factor first the logarithm $\log_{3}2(\frac{6}{p}+\frac{6}{q})$ $\implies$ $\log_{3}2(\frac{6(p+q)}{pq})$ then substitute the values of $p+q$ and $pq$ and solve to get $\log_{3}2(1.5)$ $\log_{3}3$ $=\boxed { 1}$

p+q=6,pq=24 log2(6p+6q)/pq =log2.6.6/24 =log3 =1