Solving equation with infinite exponent tower How to solve this equation for $x$ where $a>0$? The exponent tower goes on forever:
$$a=x^{x^{x^{.^{.^{.}}}}}$$
My Calculus book gives the following reasoning:
$$ln(a)=x^{x^{x^{.^{.^{.}}}}}ln(x)=a\,ln(x)$$
To conclude that: $$x=a^\frac{1}{a}$$
Why is this correct?
 A: Equation:
$$a = {x^{x^{x^{.^{.^{.}}}}}}$$
Take $ln$ of both sides, then use the power rule in right and plug $"a"$ in ${x^{x^{x^{.^{.^{.}}}}}}$:
$$ln(a) = ln({x^{x^{x^{.^{.^{.}}}}}})$$
$$ln(a) = {x^{x^{x^{.^{.^{.}}}}}} \cdot ln(x)$$
$$ln(a) = a \cdot ln(x)$$
Divide both sides by "$a"$, then rearrange left side for using power rule afterwards:
$$\frac{ln(a)}{a} = ln(x)$$
$$\frac{1}{a}ln(a) = ln(x)$$
Use power rule in left, then take $exp$ of both sides and lastly, rewrite left side as a root:
$$ln(a^\frac{1}{a}) = ln(x)$$
$$a^\frac{1}{a} = x$$
$$\sqrt [a]{a} = x$$
A: Re-write the first equation in your question as
$$a=x^a$$
and the result follows.
A: The key point is to notice that $\lim_{n\to \infty}{a_n}=\lim_{n\to \infty}{a_{n+k}}$ for any $k\in \mathbb{Z}$. It means that the limitation  does not depend on the first $k$ items. Define a sequence: $a_n=x^{a_{n-1}}$ for $n\ge 2$ and $a_n=x$ for $n=1$. 
So we can say $a=x^a$, where the $a$ in the left hand side is $\lim_{n\to \infty}{a_n}$ and the $a$ in the right hand side is $\lim_{n\to \infty}{a_{n+1}}$. 
Finally, we have $x=a^{\frac{1}{a}}$.
A: $x^{x^{x^{x^{x^\ldots}}}} = a$
This can easily be rewritten as $x^a = a$, which gives the root $x = \sqrt[a]{a}$
There is, however, one problem: $2=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}} = \sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^\ldots}}}} = 4$
A: Your book is right. Don't consider that $x^n$ is infinite. It is finite here in this case, and equal to $a$ as given in the question. Your book has a very good explanatory logic.
