>>11434636f(x) = exp(log(13)x) - exp(log(4)x) - exp(log(9)x)
f'(x) = log(13) exp(log(13)x) - log(4)exp(log(4)x) - log(9)exp(log(9)x)
notice that f'(0) <0 because log is concave. Also notice f'(1)>0 So there is a y between 0 and 1 for which f'(y)=0, and I'll let you prove that f' remains positive, which means f is monotonous for values above y and f reaches a minimum in y.
if f(1) = 0 (obvious solution), then for any z>1, f(z)>0 and for any z between y and 1 you have f(z)<0. So there are no other roots besides 1 in (y,+infinity).
As for the side (-infinity,y), we know f decreases, f(y)<0 and the limit of f at -infinity is 0. Therefore there are no roots for f on (-infinity,y).
The only remaining root is 1.
If you want a similar exercise try to find the integer solutions to x^y = y^x
