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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 $3$?2012-01-20
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    Oh, sorry. $a=1/2$ is wrong. I'll correct the question. What's wrong with $a=3?$2012-01-20
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    Sorry, forget my comment with $a=3$.2012-01-20
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    That equation is quite nice to play a bit with. Assuming that a < x it has exactly one solution where a and x are both integers, but infinitely many where they are both rational.2012-01-20
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    I don't think I understand. Do you treat $a$ as an unknown? It's a constant parameter for me.2012-01-20
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    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 $\mathbb{N}$.2012-01-20
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    @Tobias any comment as to how infinitely many solutions exist to $a^x=x^a$ when both are rational?2012-01-20
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    @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.$2012-01-20
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    cool got it now.thanks..how do you prove it btw? as in how to prove that ANY rational solution has to be of this form? it is easy to prove that any rational of this form satisfies the equation, but I am asking the other way round. You can paste the link if you please, or any hint will also suffice.2012-01-20

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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...2012-01-20
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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 $y = \mathrm{const}.$