TITLE: Solving $x^{x^{x^{x^...}}}=a$
QUESTION [5 upvotes]: Solving $$x^{x^{x^{x^...}}}=a$$
My attempt is
$$x^{x^{x^{x^...}}}\log(x)=\log (a)$$
$$a\log(x)=\log(a)$$
$$x=a^{1/a}$$
that means I can select any value of $a$ to get the root,but when I selected some values, I found them not satisfy the original equation, for example
$a=3$,$a=5$, and so on,
Is there a mistake in my procedures, 
thanks for any help

REPLY [6 votes]: Your answer is correct, at least for $x$ from $e^{-e}$ to $e^{\frac{1}{e}}$, by a proof by Euler. See this page for a much more detailed explanation, and check out the Tetration Forum online for a ton of information (much will likely be above the average mathematician (myself included) as the field is somewhat specialized, but I've always enjoyed browsing the site!)