(In reply to

Puzzle Solution by K Sengupta)

Let, log_9(x) = log_12(y) = log_16(x+y)= m(say)

-> log_10(x)/log_10(9) = log_10(y)/log_10(12) = log_10(x+y)/log_10(16) = m

-> log_10(x)= m*log_10(9), log_10(y) = m* log_10(12), and log_10(x+y) = m*log_10(16)

.......(*)

But, we observe that:

log_10(9) + log_10(16) = log_10(144) = log_10(12^2)= 2* log_10(12)

-> m*log_10(9) + m*log_10(16) = 2*m*log_10(12)

-> log_10(x) + log_10(x+y) = 2*log_10(y)

-> x(x+y) = y^2

-> y^2 -xy - x^2 = 0

-> (y/x)^2 - 2*(y/x) - 1 = 0

-> y/x = (1+/- V5)/2

But, if y/x = (1-V5)/2 ~ 0.618, then precisely one of x or y must be negative. In that situtation, log_9(x) would be undefined if x is negative, while log_12(y) would be undefined if y is negative.

Consequently, y/x = (1+ V5)/2