# A question on Exponential Equation

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.

• Do you want $2$ integer solutions, or two solutions in general, or the conditions under which you have both? – apnorton Feb 6 '14 at 2:51
• My first thought would be to break it into cases inside and outside the unit interval (maybe unnecessary). Then attempt to show the difference is never zero more than twice using the derivative and the fact that $b^x$ will dominate eventually. – David Peterson Feb 6 '14 at 2:51
• @anorton: It's always useful to assume the worst case scenario. :-) – Lucian Feb 6 '14 at 3:16
• See also: Solve $3^x + 28^x=8^x+27^x$ – Martin Sleziak May 13 '17 at 12:39

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 $f(x) = 0$.