<s>[INST] How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$ How do I solve this exponential equation?
$$5^{x}-4^{x}=3^{x}-2^{x}$$
 [/INST] $$5^x - 4^x = \int_4^5 x y^{x-1} \,dy$$
$$3^x - 2^x = \int_2^3 x y^{x-1} \,dy$$
$$= \int_4^5 x (y-2)^{x-1} \,dy$$
So the difference between $5^x - 4^x$ and $3^x - 2^x$ is
$$ \int_4^5 x (y^{x-1} - (y-2)^{x-1})\,dy$$
In the integrand here, since $y \rightarrow y^{x-1}$ is monotone whenever $x \neq 1$, the
expression $(y^{x-1} - (y-2)^{x-1})$ will either be always negative or always positive if $x \neq 1$, in which case the integral itself will be nonzero unless $x = 0$. 
We conclude that as long 
as $x \neq 0$ or $1$, $(5^x - 4^x) - (3^x - 2^x)$ is nonzero. Hence $3^x - 2^x = 5^x - 4^x$ only when $x = 0$ or $1$.
After writing all this out, I probably prefer the mean value theorem approach, but hey it's good to have more than one way of looking at a problem.
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