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Find all solutions for $3^x+6^x=4^x+5^x$

I thought about taking the limit when x goes to positive and negative infinity, but that doesn't really say anything about how many solutions there are. The derivative doesn't help at all, but I'm almost certain you need it somewhere. Does anyone have a any idea how to do this?

lulu
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רואי בורלא
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    Have you found any solutions? Have you tried to graph the difference (i.e. $3^x+6^x-(4^x+5^x)$? That's often a good way to count the solution.... – lulu Jan 18 '24 at 11:37
  • Yes, I even graphed it on desmos after struggling for a bit, but I need to prove it formally. – רואי בורלא Jan 18 '24 at 11:42
  • You can trivially check that $x=0$ and $x=1$ are solutions. – PierreCarre Jan 18 '24 at 11:46
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    By generalized descrate's rule of sign, the number of solution is at most 2. Since you already have two trivial solutions, $x = 0$ or $1$, you have all of them. For the actual statement and refs, see my answer to a similar question ten years ago. – achille hui Jan 18 '24 at 12:09
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    I didn't know about generalized DRS, this is new to me, powerful indeed. – zwim Jan 18 '24 at 12:13
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    @zwim I have never heard of that either. – Peter Jan 18 '24 at 12:21
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    @achillehui Doesn't descarte's rule of signs only apply to polynomials? – fleablood Jan 18 '24 at 16:25
  • @fleablood it is a result by Laguerre which extends the rule to "polynomials" whose exponents are any real number. This in turn is equivalent to sum of exponential functions we see here. see ref in the link I mentioned before. – achille hui Jan 18 '24 at 21:23

2 Answers2

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Rewrite the given equation as $$\frac{6^x-5^x}{6-5}=\frac{4^x-3^x}{4-3}$$ This like applying LMVT to $f(t)=t^x$ in $(5,6)$ and $(3,4)$. then $$\implies xt_1^{x-1}=xt_2^{x-1}, t_1\ne t_2, t_1\in (5,6), t_2\in (3,4).$$ We get two solutions $x=0,1$.

Z Ahmed
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If you want to emprically check, you can see after $x=1$ that $\frac{5^x+4^x}{6^x+3^x}$ converges slowly to zero. The expected growth would indicate that as a solution is near the expression also nears to $1$. You can check then since it is $< 1$ that the limits as $$\lim_{x\rightarrow \infty}\frac{5^x+4^x}{6^x+3^x} = 0$$ $$\lim_{x\rightarrow -\infty}\frac{5^x+4^x}{6^x+3^x} = 0$$ and hence there are no more solutions other than $x=0$ and $x=1 $...

Mako
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