Could you please explain me how to solve:
If $p\:=\:\log _{12}\left(18\right)$ and $q\:=\:\log _{24}\left(54\right)$, $pq\:+\:5\left(p-q\right)\:=\:1$
I tried this way:
$p = \frac{2\log\left(3\right)+\log\left(2\right)}{2\log\left(2\right)+\log\left(3\right)}$, $q = \frac{3\log\left(3\right)+\log\left(2\right)}{3\log\left(2\right)+\log\left(3\right)}$
But not sure what to do next? I'd be grateful if you can help me!
HINT:
The idea is to eliminate $\log2, \log3$
$$(2p-1)\log2=(2-p)\log3$$
$$(3q-1)\log2=(3-q)\log3$$
Divide the the first relation by the second and rearrange.