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.
Best Answer
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. ;-)