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.
Here's a partial answer.
Let $f(x) = a^x + b^x - c^x - d^x$. We have $f(0) = 0$ and $f'(0) = \log a + \log b - \log c - \log d = \log\left( \dfrac{ab}{cd} \right)$. As $x \to +\infty$, $b^x$ dominates, while as $x \to -\infty$, $a^x$ dominates, so $f(x)$ is positive in both cases. If $ab \ne cd$, $f'(0) \ne 0$ and by the Intermediate Value Theorem there will be at least one more solution of $f(x) = 0$.