TITLE: How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$
QUESTION [10 upvotes]: How do I solve this exponential equation?
$$5^{x}-4^{x}=3^{x}-2^{x}$$

REPLY [3 votes]: Let $f:[0.25,2] \to \mathbb R \,;\, f(a)=(3.5-a)^x+(3.5+a)^x \,.$
Then $f'(a)=x[(3.5+a)^{x-1}-(3.5-a)^{x-1}]$.
Claim: If $x \notin \{ 0,1 \}$ we have $f'(a) \neq 0 \forall a $.
Indeed, in this case $f'(a) =0 \Rightarrow (3.5+a)^{x-1}=(3.5-a)^{x-1} \Rightarrow a=0$ which is not in our domain.
This proves that for $x \neq 0,1$, $f$ is one to one on our domain, and hence
$$f(0.5) \neq f(1.5) \Rightarrow 3^x+4^x \neq 2^x+5^x $$