smc

We can rewrite the equation as $\frac{\log p}{\log 9}=\frac{\log q}{\log 12}=\frac{\log (p+q)}{\log 16}$. Then, the system can be split into 3 pairs: $\frac{\log p}{\log 9}=\frac{\log q}{\log 12}$, $\frac{\log q}{\log 12}=\frac{\log (p+q)}{\log 16}$, and $\frac{\log p}{\log 9}=\frac{\log (p+q)}{\log 16}$. Cross-multiplying in the first two, we obtain: $$(\log 12)(\log p)=(2\log 3)(\log q)$$ and $$(\log 12)(\log (p+q))=(2\log 4)(\log q)$$ Adding these equations results in: $$(\log 12)(\log p(p+q))=(2\log 12)(\log q)$$ which simplifies to $$p(p+q)=q^2$$ Dividing by $pq$ on both sides gives: $\frac{p+q}{q}=\frac{q}{p}=\frac{p}{q}+1$. We set the desired value, $q/p$ to $x$ and substitute it into our equation: $\frac{1}{x}+1=x⟹x^2-x-1=0$ which is solved to get our answer: $\boxed{\frac{1+\sqrt{5}}{2}}$.