Solving $x^{x^{x^{x^...}}}=a$

Question

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

Somewhat close to: http://math.stackexchange.com/questions/1581868/an-infinitely-powered-expression-large-sqrt2-sqrt2-sqrt2-cdot — NoChance, Dec 27 '15 at 21:11
I think there are a dozen questions almost exactly like this already on the site. — 2'5 9'2, Dec 27 '15 at 21:11
Read about infinite tetrations. In particular, your left-had expression only has meaning for $x\in\left[e^{-e},e^{1/e}\right]\approx\left[0.065988\ldots,1.4476\right]$. — 2'5 9'2, Dec 27 '15 at 21:13

score 6 · Accepted Answer · answered Dec 27 '15 at 21:21

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!)

Batominovski · Answer 2 · 2015-12-27T21:19:32.973

0

If $x\in\mathbb{R}_{>0}$ is such that $x^{x^{x^{\cdot\,^{\cdot\,^\cdot}}}}$ has a limit $a$. Then, $x^a=a$.

edited Dec 27 '15 at 21:19

answered Dec 27 '15 at 21:07

Batominovski

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Are you saying that another approach shows that the OP's approach was correct? The point of this answer is not clear. – Rory Daulton Dec 27 '15 at 23:59
Yes, that is my point. – Batominovski Dec 28 '15 at 00:42

Solving $x^{x^{x^{x^...}}}=a$

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