I came across the following question a few week ago (Exponential equation+derivative):

Solve $3^x+28^x=8^x+27^x$.

The answer for the above question is 0 and 2.

I generalized the question, as follows.

Suppose $0<a<c<d<b$. Show that the number of solutions to the equation $$a^x+b^x=c^x+d^x$$

(i) is 1 if $ab=cd$

(ii) is 2 if $ab \neq cd$.

I proved (i) but I have not idea how to prove (ii).

I verify (ii) by ploting lots of graphs using WolframAlpha.

Any hints?

Note: from the post (How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$), we also know that the solution of $a^x+b^x=c^x+d^x$ is 0 and 1 if $b-d=c-a=1$. I wonder what conditions will make the equation has 2 integer solutions.