Using Rolle's Theorem to show that $3^x+4^x=5^x$ iff $x=2$ I need to use Rolle's Theorem to show that the only real solution to $3^x+4^x=5^x$ is $x=2$. Here's what I have:
Proof: Note that a number $x$ satisfies $3^x+4^x=5^x$ if and only if $f(x)=0$ where $f(x)=3^x+4^x-5^x$. Obviously $x=2$ is solution since $f(2)=0$. Suppose that there exists a second solution $x_2$. By Rolle's Theorem, there exists a number $c$ between $2$ and $x_2$ such that $f'(c)=0$. Note that $f'(x)=3^x\ln(3)+4^x\ln(4)-5^x\ln(5)$ and that $f'(1.287)=0$.
Rolle's Theorem doesn't help us here since there is a value where the derivative is zero - so I can't get a contradiction. That is, there could be another solution that is less than 1.287.
Could someone maybe point me in the right direction here? Am I approaching the proof correctly?

What about the function $(3/5)^x+(4/5)^x-1$. The derivative of this does not have a root. 
