Xueqin!

log₉ x = log₁₂ y
ln x / ln 9 = ln y / ln 12
ln x = ln y * (ln 9 / ln 12)
x = y^{ln 9 / ln 12}

Replace in the other one, and you should get somewhere.

Can you explain how you got from 3rd to 4th step?(oops accidentally edited your post)

Originally Posted by aaaaaa123456789

Can you explain how you got from 3rd to 4th step?(oops accidentally edited your post)

ln x = ln y * (ln 9 / ln 12)
ln x = ln y^(ln 9 / ln 12)
x = y ^ (ln 9 / ln 12)

Can you explain further? I still can't find X/Y.

typing answer now, one sec

EDIT:

ln(x)/ln(9) = ln(y)/ln(12) = ln(x+y)/ln(16)

let x = ay
a = x/y we are trying to find that

(ln(a) + ln(y)) / ln(9) = ln(y) / ln(12) = (ln(y) + ln(a+1)) / ln(16)

ln(y) + ln(a) = ln(y) * ln(9) / ln(12)
ln(y) + ln(a+1) = ln(y) * ln(16) / ln(12)

add those two equations:

2 * ln(y) + ln(a) + ln(a+1) = ln(y) * ((ln(9) + ln(16)) / ln(12))
2 * ln(y) + ln(a) + ln(a+1) = 2 * ln(y)
ln(a) + ln(a+1) = 0
ln(a * (a+1)) = 0
a^2 + a = 1
(a+1/2)^2 = 5/4

a = -1/2 ± √(5/4) = (± √5 - 1) / 2
x/y = (± √5 - 1) / 2