# Difference between revisions of "1988 AHSME Problems/Problem 26"

## Problem

Suppose that $p$ and $q$ are positive numbers for which

$\log_{9}(p) = \log_{12}(q) = \log_{16}(p+q)$

What is the value of $\frac{q}{p}$?

$\textbf{(A)}\ \frac{4}{3}\qquad \textbf{(B)}\ \frac{1+\sqrt{3}}{2}\qquad \textbf{(C)}\ \frac{8}{5}\qquad \textbf{(D)}\ \frac{1+\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{16}{9}$

## Solution

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: $\frac{1}{x} + 1 = x \implies x^2 - x - 1 = 0$ which is solved to get our answer: $\boxed{\text{(C) } \frac{1 + \sqrt{5}}{2}}$. -lucasxia01