Question 415790
If {{{x^y = 16}}}, there are only a few possible solutions for x and y.
16^1 = 16, but the problem states that x < y.
4^2 = 16, but again x < y.
So the only possibility is 2^4 = 16, so x=2, y=4 -> x-y = 2-4 = -2.