Put:
$\displaystyle u=\log_9(x)=\log_{12}(y)=\log_{16}(x+y)$
Then:
$\displaystyle 9^u=x,\ 12^u=y,\ 16^u=x+y$
So:
$\displaystyle
\frac{x+y}{y}=\left[\frac{16}{12}\right]^u=\left[\frac{4}{3}\right]^u
$
and:
$\displaystyle
\frac{y}{x}=\left[\frac{12}{9}\right]^u=\left[\frac{4}{3}\right]^u
$
Therefore if $\displaystyle z=x/y$, we have:
$\displaystyle
z+1=1/z
$
which is a quadratic in $\displaystyle z$, one (or more of which's solutions)
should solve the problem.
RonL
RonL