5
$\begingroup$

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.

  • 0
    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

2 Answers 2

4

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...

  • 0
    Wow, thanks! I have no idea how I missed that...2012-01-20
0

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 $y = \mathrm{const}.$