Charlie, I think you set the problem up correctly but forgot some of the rules of logs.

My method here is long but (I think) correct.  Someone please show me an easier way.

let n be equal to the 3 logarithms

and re-writing in exponental form:

x=9^n     y=12^n      x+y=16^n

substituting the first two into the third:

9^n+12^n=16^n or

[(4/3)^2*9]^n  =  9^n  +  (4/3)^n*9^n

divide both sides by 9^n and rearranging:

[(4/3)^n]^2  -  (4/3)^n  =  1

by "completing the square":

[(4/3)^n]^2  -  (4/3)^n  +  1/4  =  1  +  1/4  or:

[(4/3)^n - 1/2] ^ 2 = 5/4

taking the square root of both sides:

(4/3)^n - 1/2 = +/- ©÷¡î5 and solving for n:

n=ln[(1 +/- ©÷¡î5)/2] / ln [4 / 3]

this is equivelent to taking the log base ((4/3) of the quantity

[(1 +/- ©÷¡î5)/2]

Now back to the question (find y/x)

y / x = 12^n / 9^n = (4/3)^n*9^n / 9^n =(4/3)^n

now y/x is (4/3) raised to the log(base 4/3) of a quantity therefore only the quantity is left

y/x = (1 +/- ©÷¡î5)/2

or the golden ratio.