Any neat proof that $0$ is the unique solution of the equation $4^x+9^x+25^x=6^x+10^x+15^x$? It is obvious that both $f(x)= 4^x+9^x+25^x$ and $g(x)=6^x+10^x+15^x$ are strictly monotonic increasing functions. It is also easy to check that $0$ is a solution of the equation.
Also I chart the functions, and it looks that for any $x$, $f(x)>g(x),$ which can be somehow proof by studying the derivative of the $h(x)=f(x)-g(x)$ and showing that $(0,0)$ is an absolute minimum point for $h(x).$
However $h(x)$ is a function with a messy derivative, and is not looking easy (for me) to find the zeroes of this derivative.
Does anyone know an elegant proof (maybe an elementary one, without derivatives) for this problem?

I guess you want a proof that works for the continuous case. If we would know that $x$ is always an integer, then we could say that $f(x)$ is an even number for any $x$ integer except $0$, and $g(x)$ is odd for any $x$ integer except $0$. So, $x$ must be zero. 
