Multi-Logarithms

$\log_9{(x)} = log_{12}{(y)} = log_{16}{(x+y)}\\\implies~\dfrac{logX}{log9} =\dfrac{logY}{log12}=\dfrac{log(X+Y)}{log16}=\dfrac{log\dfrac{Y}{X}}{log\dfrac{12}{9}}=a..say\\\implies~X=9^a=3^{2a},..Y=12^a=3^a*4^a,..X+Y=16^a=4^{2a}\\ and~~\dfrac{Y}{X}=(\dfrac{4}{3})^a.......(1)\\\therefore~ 3^{2a}+ 3^a*4^a=X+Y=4^{2a}~~~\implies~(\dfrac{4}{3})^{2a}- (\dfrac{4}{3})^a -1=0.\\We ~get~a~quadratic~in~(\dfrac{4}{3})^a~~and~ solving ~ and ~noting~that ~\\ we~ are ~dealing~ with ~log~ that ~has~ domine~ >0,\\(\dfrac{4}{3})^a= \dfrac{1 + \sqrt{5}}{2}.~~From~ (1)\therefore~\dfrac{1 + \sqrt{5}}{2}=\dfrac{Y}{X}$ $\boxed{\huge{ \dfrac{1 + \sqrt{5}}{2}}}$