# 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 $$1.$$
Since $$f(a)=a^x$$ is concave for $$x\in[0,1]$$ and convex for $$x\not\in(0,1)$$, the definitions of concavity and convexity say \begin{align} \color{#C00}{3^x}+\color{#090}{4^x} &=\color{#C00}{\left(\frac23\cdot2+\frac13\cdot5\right)^x}+\color{#090}{\left(\frac13\cdot2+\frac23\cdot5\right)^x}\tag1\\[3pt] &{\ge\atop\le}\color{#C00}{\left(\frac23\cdot2^x+\frac13\cdot5^x\right)}+\color{#090}{\left(\frac13\cdot2^x+\frac23\cdot5^x\right)}{{\quad\text{if }x\in[0,1]}\atop{\quad\text{if }x\not\in(0,1)}}\tag2\\[6pt] &=2^x+5^x\tag3 \end{align} Explanation:
$$(1)$$: write $$3$$ and $$4$$ as convex combinations of $$2$$ and $$5$$
$$(2)$$: definition of $$\text{concavity}\atop\text{convexity}$$
$$(3)$$: combine like terms
Furthermore, equality holds only if $$f$$ is linear on $$\{2,3,4,5\}$$, and that happens iff $$x\in\{0,1\}$$.