Contest Math Question on Logarithms

Question

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.

Answer 1

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?

Answer 2

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 $d$.

Answer 3

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. ;-)

Answer 4

Let $$ \log _9 p=\log _{12} q=\log _{16}(p+q)=k $$ then $$ 9^k+12^k=16^k \Leftrightarrow \left(\frac{3}{4}\right)^k+1=\left(\frac{4}{3}\right)^k \Leftrightarrow \frac{p}{q}+1=\frac{q}{p} \Leftrightarrow \frac{q}{p} =\frac{1+\sqrt{5}}{2} $$