system
Given the following post, respond as if you're commenting on Reddit:
human
Solutions to the equation a^a = b^b?

After seeing the thread about the possible solutions to a^b = b^a, I was wondering what will happen if we were to look for the solutions to a^a = b^b where a =/= b.

One possible solution is when a = 1/2 and b = 1/4. Intuitively, the possible solutions for a and b should be in the interval (0,1) since if a > b > 1, then a^a > b^b.

By considering the function y = x^x, it can be shown that it achieves a global minimum at 1/e and that the graph concaves up for every x in its domain. So, in the interval (0, 1), there ought to be infinite solutions as horizontal straight lines can be drawn to cut the graph of y twice in this interval. (Please do correct me if this reasoning is wrong.)

So, the thing is, is it possible to obtain all possible pairs of (a, b) that are solutions to a^a = b^b ?
gpt
You can do the same trick. Firstly, taking logs turns this into aloga=blogb, which is simpler to work with. Set b=a^(r) and plug it in to get aloga=ra^(r)loga, or a^(1-r)=r, which means a=r^(1/[1-r]), which gives b=r^(r/[1-r]). Any 0<a<1 can be written uniquely in this way and any r>0 gives such an a. And similarly for b. So this gives all solutions.

If you want rational solutions, we'll need 1/(1-r)=N, an integer, which means r=1-1/N. In this case, a=(1-1/N)^(N) and b=(1-1/N)^(N-1). These are the solutions when a and b are both rational. Eg: a=1/4, b=1/2.