Re: Solve y^x = x^y ?

From: David W. Cantrell (DWCantrell_at_sigmaxi.org)
Date: 11/11/04 

Date: 11 Nov 2004 22:22:48 GMT

Ioannis <morpheus@olympus.mons> wrote:
> David W. Cantrell wrote:
> [snip]
> >>I *think* that when BOTH x and y are > 0, the only solutions seem to me
> >>to be the ones exhibited by the Lemma above, so it was a
> >>misunderstanding of the term "all".
> >
> > Does the lemma cover the trivial solutions: x = y ?
>
> Obviously not.

I wasn't being flippant when I asked that, BTW. The trivial solutions when
both x and y are positive can be written in terms of the Lambert W
function, as I did in my article.

When 0 < x <= e, y = LambertW(0, -log(x)/x)/(-log(x)/x)

simplifies to y = x, as does

y = LambertW(-1, -log(x)/x)/(-log(x)/x) when x >= e.

So, not remembering your lemma well when I asked the question above,
I wasn't 100% certain that it didn't also yield those trivial solutions.

David

> > In the meantime, you could use Google groups:
> > <http://groups.google.com/groups?selm=11fdb874.5ee5dbe7%40usw-ex0104-08
> > 7.remarq.com>

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