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Find the values of $x$ where $x \in R $ which satisfy the equation: $2^x+5^x=3^x+4^x$ other than $x=0, 1.$

I got the answer using Mean Value Theorem that there is no other solution but is there any other method to do it? (Is this a one-way math?)

I thought of proving that $5^x> 3^x+4^x$ $\forall$ $x>2$ but we have $x \in R $ and hence it is incomplete. The last digit calculation also doesn't work out since $x \in R $.

Would be glad if someone could suggest some other method to do it.

Tapi
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    Hint: Prove that derivative of LHS is bigger than derivative of RHS for $x\ge1$ then show that there is no more solution. – Taha Direk Jul 02 '19 at 15:28
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    You may see: https://math.stackexchange.com/questions/3275626/number-of-real-roots-of-3x4x-2x5x-with-proof/3275633#3275633 – Z Ahmed Jul 02 '19 at 15:36

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This equation can be written as $$ 5^x - 4^x = 3^x - 2^x$$ Let $f(a,x) = (a+1)^x - a^x$. We can write our equation as $$ f(4,x) = f(2,x) $$ $$ \int_2^4 \frac{\partial f(a,x)}{\partial a} da = 0$$ We have $$ \frac{\partial f}{\partial a} = x \big((a+1)^{x-1} - a^{x-1}\big)$$ For $x>1$ or $x<0$ we have $\frac{\partial f}{\partial a}>0$ for $a\in(2,4)$, so the integral cannot be equal to $0$. Similarily, for $0<x<1$ we have $\frac{\partial f}{\partial a}<0$, and again the integral cannot be equal to $0$. Cases $x=0,1$ remain, for which we can find that the equation is satisfied.

Adam Latosiński
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