>Forgive the typos please, and sorry
>about the subject line.

Do not worry about that. I have pointed this out because the two problems are in fact very different. True, they lead to the same kind of solutions and both deal with repeated exponents. However, the problem I discussed in the text, has solutions for any (positive) right hand side.

>Fooling around with the generalization,
>f(x)=a, with the solution x=a^(1/a),
>I tried a=4. Which would
>have the solution x=4^1/4 =
>2^1/2. i.e. f(x)= 2 and
>f(x)= 4 have the same
>solution. f(2^(1/2))=2 and f(4^(1/4)) =
>4, then 2=4 and
>I'm the pope, and everything
>else is true.

This is a curious observation indeed. And I gather you are not the Pope. But again, your remark underscores the difference between the two problems. Two eqautions xx2 = 2 and xx4 = 4 have the same solution 21/2 = 41/4 without engendering a contrudiction.

> Actually,
>the problem only works for
>values of a between 1/e
>and e. (|ln(a)|<1.) And the
>infinite nested series only converges
>for values of x between
>1/e^e and e^1/e.

Yes, this is true. I recollect seeing this discussed in some book, but can't now place it. Do you have a reference?

> I
>always thought it was a
>fun problem.

Absolutely.

Alexander Bogomolny