Post #2324823

@divbyzero@mathstodon.xyz I was surprised by this statement, and then I read more carefully and saw that this is not f(x) = exp(x) - log(x), but rather f(x, y) = exp(x) - log(y), and at that point it's totally obvious. By setting y = 1 you recover the exponential function. By setting x = 0 you get 1 - log(y), which you can play around with a bit to extract log(y). And then you have exponentials and logs and subtraction and that's obviously enough to give you everything.