Number of real roots of $3^x+4^x=2^x+5^x$ with proof

Question

This equation $$3^x+4^x=2^x+5^x$$ has two obvious real roots. The question is if it has more real roots than two. A proof is required in any case.

Answer 1

Let $f(t)=t^x.$

1. $x>1$ or $x<0$.

Since $f$ is a convex non-linear function and $(5,2)\succ(4,3),$ by Karamata we obtain: $$f(5)+f(2)>f(4)+f(3)$$ or $$5^x+2^x>4^x+3^x,$$ which says that in this case our equation has no roots.

2. $0<x<1.$

Here, $f$ is a concave function and by Karamata again we obtain: $$5^x+2^x<4^x+3^x,$$ which says that in this case our equation has no roots.

But $1$ and $0$ are roots, which says that our equation has two roots exactly.

Answer 2

We consider the equation $$3^x+4^x=2^x+5^x~~~~(1)$$ Use Lagranges Mean Value Theorem (LMVT) for the function $f(t)=t^x$ for two intervals $(2,3)$ and $(4,5)$. So $$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, ~~~t_1 \in (2,3)~~~~(2)$$ and $$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, ~~~t_2 \in (4,5)~~~~(3).$$ By equating (2) and (3), we get (1) and $$xt_1^{x-1} = xt_2^{x-1}, ~~~t_1 \ne t_2\Rightarrow x=0~ \mbox{or}~ x=1.$$ Hence Eq. (1) can have only two real roots.