How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$

Question

How do I solve this exponential equation?
$$5^{x}-4^{x}=3^{x}-2^{x}$$

$x=0,1$ seems to be the only real solution. for $x\gt 1$, the L.H.S. growth rate is more than that of R.H.S. and for $x\lt 0$, L.H.S. $\lt $ R.H.S — Aang, Feb 02 '13 at 04:09
No, it is a religious prohibition. Well, it will be, once I have finished designing the religion. — Will Jagy, Feb 02 '13 at 04:10
you should add whether x is from he integers, reals or complex. If it is from the integers then Euler's totient gives also an answer, optimally if you rewrite $5^x-3^x=2^x(2^x-1)$ — Gottfried Helms, Feb 02 '13 at 08:25

score 39 · Accepted Answer · edited Aug 21 '15 at 23:28

39

Hint:

$$\frac{5^x-4^x}{5-4} = \frac{3^x-2^x}{3-2}$$

Now use the Lagrange Mean Value Theorem over the intervals $(2,3)$ and $(4,5)$. The equality holds only when $x \in \{0,1\}$.

edited Aug 21 '15 at 23:28

Fabrosi

673

answered Feb 02 '13 at 04:19

juantheron

53,015

5

@BunsOfWrath I think this was a pointless edit on a very highly-upvoted answer that's more than two years old. – pjs36 Aug 21 '15 at 23:33
3

@pjs36 Yeah, maybe. It wasn't immediately obvious to me that $(2,3)$ and $(4,5)$ were intervals, so I edited for that. The other changes were just because the TeX was a bit sloppy. – Fabrosi Aug 21 '15 at 23:51

score 16 · Answer 2 · answered Feb 02 '13 at 04:51

$$5^x - 4^x = \int_4^5 x y^{x-1} \,dy$$ $$3^x - 2^x = \int_2^3 x y^{x-1} \,dy$$ $$= \int_4^5 x (y-2)^{x-1} \,dy$$ So the difference between $5^x - 4^x$ and $3^x - 2^x$ is $$ \int_4^5 x (y^{x-1} - (y-2)^{x-1})\,dy$$ In the integrand here, since $y \rightarrow y^{x-1}$ is monotone whenever $x \neq 1$, the expression $(y^{x-1} - (y-2)^{x-1})$ will either be always negative or always positive if $x \neq 1$, in which case the integral itself will be nonzero unless $x = 0$.

We conclude that as long as $x \neq 0$ or $1$, $(5^x - 4^x) - (3^x - 2^x)$ is nonzero. Hence $3^x - 2^x = 5^x - 4^x$ only when $x = 0$ or $1$.

After writing all this out, I probably prefer the mean value theorem approach, but hey it's good to have more than one way of looking at a problem.

score 3 · Answer 3 · answered Feb 02 '13 at 05:28

Let $f:[0.25,2] \to \mathbb R \,;\, f(a)=(3.5-a)^x+(3.5+a)^x \,.$

Then $f'(a)=x[(3.5+a)^{x-1}-(3.5-a)^{x-1}]$.

Claim: If $x \notin \{ 0,1 \}$ we have $f'(a) \neq 0 \forall a $.

Indeed, in this case $f'(a) =0 \Rightarrow (3.5+a)^{x-1}=(3.5-a)^{x-1} \Rightarrow a=0$ which is not in our domain.

This proves that for $x \neq 0,1$, $f$ is one to one on our domain, and hence

$$f(0.5) \neq f(1.5) \Rightarrow 3^x+4^x \neq 2^x+5^x $$

score 1 · Answer 4 · answered Feb 02 '13 at 04:11

Only method I can think of is trying graphically. Since their slopes vary a lot after $(1,1)$ and before $(0,0)$, all possible intersections can lie between these two points.

How to find all those possible intersections is something I'd need to work out.

How do I solve this exponential equation? $5^{x}-4^{x}=3^{x}-2^{x}$

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