Contest Math Question on Logarithms I am trying to solve a question from the AoPS Vol. 2 book.
The question is as follows:

Suppose the $p$ and $q$ are positive numbers for which:
$$\log_9 p = \log_{12}q = \log_{16}(p+q)$$
What is the value of $q/p$? (AHSME 1988)

I only have knowledge of a few of the basic log identities/properties to work with, but they are not getting me anywhere, I find myself only coming up with some obvious equalities that do not lead me closer.
One thing I tried was to rewrite the equalities as:
$$\frac{\log p}{\log 9} = \frac{\log{q}}{\log 12} = \frac{\log(p+q)}{\log 16}$$
where $\log = \log_{10}$.
Then, rearranging the first equality shows:
$$\log_q p = \log_{12} 9$$
I thought at first that I could just take $q=12$ and $p=9$ directly via substitution, but substituting in these values does not satisfy the equality given in the problem statement since $\log_{16} 21 \neq 1$. (What's actually going on here by the way? Is it because you can have multiple values of $p$ and $q$ besides $12$ and $9$ that satisfy the equality?)
I feel that I am missing some key step in my approach to be able to solve this that is beyond just manipulating the formulas/identities. I am wondering if someone can point me in the right direction about how to approach this, and perhaps explain how one would reach such an approach.
Thanks.
 A: Let $x$ denote this quantity that all three expressions are equal to. Then
$$
9^x = p, 
\quad 
12^x = q, 
\quad\text{and}\quad 
16^x = p + q. 
$$
From the first two equations,
$$
\frac{q}{p} = \frac{12^x}{9^x} 
= \biggl( \frac{4}{3} \biggr)^x, 
$$
and from the first and third equation,
$$
1 + \frac{q}{p} = \frac{p + q}{p}
= \frac{16^x}{9^x} 
= \biggl( \frac{16}{9} \biggr)^x 
= \biggl( \frac{4}{3} \biggr)^{2x}. 
$$
Letting $r = \frac{q}{p}$, and putting these two equations together,
$$
1 + r = r^2, 
$$
a golden equation which I'm sure you can solve. ;-)
A: Here is another approach:
Once you have gotten to $\displaystyle\frac{\log p}{\log 9} = \frac{\log{q}}{\log 12} = \frac{\log(p+q)}{\log 16}$, you may cross multiply the first equality to get $(\log{12})(\log p)=(2\log3)(\log q)$.
Cross-multiplying the second inequality gives $(\log12)(\log(p+q))=(\log q)(2\log 4)$.
Adding these two gives $$(\log12)(\log(p(p+q)))=(\log q)(2\log12)\implies p(p+q)=q^2\implies\dfrac{p+q}{q}=\dfrac qp.$$ Can you take it from here?
A: Here's a hint: if $a/c = b/d$, then show these equal the quotient gotten by dividing the average of $a$ and $b$ by the average of $c$ and $d$.
A: Let $$
\log _9 p=\log _{12} q=\log _{16}(p+q)=k
$$
then $$
9^k+12^k=16^k \Leftrightarrow \left(\frac{3}{4}\right)^k+1=\left(\frac{4}{3}\right)^k \Leftrightarrow \frac{p}{q}+1=\frac{q}{p} \Leftrightarrow 
\frac{q}{p} =\frac{1+\sqrt{5}}{2}
$$
