given that ${\log_9 p} = {\log_{12} q} = \log_{16}(p+q)$ find the value of $q/p$

Question

This is not homework, it's just a brain teaser which I can't solve, just some hints should be sufficient, I know that from this we get: $$ (1/4)\log_2(p+q) = (1/2)\log_3 p = \frac{\log_3 q}{1+2\log_3 2} $$ now I'd like to combine these quantities in some way so I can see the value of $q/p$ from this, but I can't seem to be able to figure it out.

Answer 1

So there is a number $x$ such that $9^{x}=p$, $12^x=q$ and $16^x=p+q$
Therefore $9^x + 12^x = 16^x$
One final hint: $12=3\times 4$

Answer 2

Let $L$ denote the common logarithmic value. Thus $p=9^L$, $q=12^L$, and $p+q=16^L$. Let $r=q/p=12^L/9^L=(4/3)^L$. Then

$$r^2=(16/9)^L=(p+q)/p=1+r$$

Can you take it from there?