Question

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 $q/p$?

Answer 1

Given $p,q>0$ and $\log_9p=\log_{12}q=\log_{16}(p+q)=x$ (say).

Then,

$p=9^x\\ q=12^x\\ p+q=16^x$

From the above 3 equations,

$\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}$

From the equations, we get

$\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}$

Since $p,q>0$, the negative value is rejected.

$\frac{q}{p}=\frac{1 + \sqrt{5}}{2}$