# 1988 AHSME Problems/Problem 26

## Problem

Suppose that $p$ and $q$ are positive numbers for which $$\operatorname{log}_{9}(p) = \operatorname{log}_{12}(q) = \operatorname{log}_{16}(p+q).$$ 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 1

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: $\boxed{\text{(D) } \frac{1 + \sqrt{5}}{2}}$. -lucasxia01

## Solution 2

For some number t: $p = 9^{t}$ $q = 12^{t}$ $p + q = 16^{t}$

Next we can divide $p + q$ by $p$ to obtain $\frac{p+q}{p} = 1 + \frac{q}{p}$

Furthermore, we know that $\frac{p+q}{p} = (\frac{16}{9})^{t}$ and $\frac{q}{p} = (\frac{4}{3})^{t}$

Substituting into the previous equation, we get $(\frac{16}{9})^{t} = 1 + (\frac{4}{3})^{t}$

Let $x = (\frac{4}{3})^{t}$ and we can observe that $x = \frac{q}{p}$, then similarly to solution 1: $x^2 = 1 + x$, in which we get: $\boxed{\text{(D) } \frac{1 + \sqrt{5}}{2}}$ - ehmmaq

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 