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.
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Michael Rozenberg
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Z Ahmed
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1By Laguerre's "Generalized Descartes' rule of signs", the equation has at most 2 roots. Since you find two real roots. It is all it has. For a proof, see refs in my answer many years ago. – achille hui Jun 27 '19 at 04:00
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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.
Z Ahmed
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your answer makes sense, but why you are comparing slopes in the last equation? – xrfxlp Jun 27 '19 at 04:12
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@AjayMishra They are really just differences, since denominators are $1.$ – coffeemath Jun 27 '19 at 05:44
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@Nochance Thanks, though this question was declared Duplicate, yet people took an interest in its present solutions. – Z Ahmed Jul 03 '19 at 05:17
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Let $f(t)=t^x.$
- $x>1$ 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.
- $0<x<1.$
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.
Michael Rozenberg
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