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Let $a\in (1,e)\cup(e,\infty).$ I'd like to show that the equation $a^x=x^a$ has exactly two positive solutions, and one is larger and one smaller than $e.$ Is it even possible to show? I think I've tried everything.

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What if $a=1/2$? Or $3$? – David Mitra Jan 20 '12 at 12:09
Oh, sorry. $a=1/2$ is wrong. I'll correct the question. What's wrong with $a=3?$ – user23211 Jan 20 '12 at 12:14
Sorry, forget my comment with $a=3$. – David Mitra Jan 20 '12 at 12:21
let $a,x\epsilon \mathbb{N}$.Then, either $x|a$ or $a|x$. Assume $a|x$. Then we have $x=ma$, where $m\epsilon \mathbb{N}$.Given $a^x=x^a$. Thus, we have $a^{ma}=(ma)^a$ i.e. $ a^{ma}=m^a\times a^a$ i.e. $a^{(m-1)a}=m^a$ i.e. $(a^{m-1})^a=m^a$, hence we have $a^{m-1}=m$. Thus, the solution of original equation boils down to the solution of $$a^{m-1}=m\cdots(1)$$Now, we know that $2^{m-1}>m$ for $\forall m>2.$ Hence, $\forall a,m>2,$ equation $(1)$ can't be satisfied. Above equation is satisfied in $\mathbb{N}$ only when $a,m=(2,2)$. This leads to the only solution $a=2, x=4$ in $\mathbb{N}$. – NikBels Jan 20 '12 at 13:07
@Nikhil Bellarykar Of course it's true for any positive rational $a=x$. Other than that, the set of rational solutions is $a=\left(1+\frac{1}{n}\right)^n,\;y=\left(1+\frac{1}{n}\right)^{n+1}$ for $n=1,2,...$ when $x>a.$ – user23211 Jan 20 '12 at 13:24
up vote 4 down vote accepted

The equation is equivalent to $\frac{\log(a)}{a} = \frac{\log(x)}{x}$. Now look at the value of the left hand side and the graph of the right hand side...

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Wow, thanks! I have no idea how I missed that... – user23211 Jan 20 '12 at 12:44

For positive numbers, your equation is equivalent to $\sqrt[a]{a} = \sqrt[x]{x}$, so you have to consider the graph of the function $$ y = \sqrt[x]{x} $$ with sections of the form $y = \mathrm{const}.$

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