# Solving an exponential equation with different bases

Solve the equation $2^x + 5^x = 3^x + 4^x$. I can figure out two special solutions $x=0$ and $x=1$, and I try to prove that they are the only two solutions. However, I find it hard to do so because I can't prove the monotony given there are also exponential in the derivative. Any hints to that?

$$2^x+5^x=3^x+4^x \iff5^x-4^x=3^x-2^x$$ Lagrange's theorem applies (intermediate value theorem) functions: $$f:[4, 5]\to \mathbb R,\ g:[2, 3]\to \mathbb R,\ f(u)= u^x,\ g(v)=v^x$$ Exist $$c\in [4, 5]$$ and exist $$d\in [2, 3]$$ so $$5^x-4^x = xc^{x-1}$$ and $$3^x-2^x=xd^{x-1}$$.
The equation is written as equivalent: $$xc^{x-1}= xd^{x-1}.$$ It follows that equation solutions are $$0$$ and $$1.$$