Solving equation with infinite exponent tower

Question

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?

Answer 1

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.

Answer 2

Re-write the first equation in your question as

$$a=x^a$$

and the result follows.

Answer 3

$x^{x^{x^{x^{x^\ldots}}}} = a$

This can easily be rewritten as $x^a = a$, which gives the root $x = \sqrt[a]{a}$

There is, however, one problem: $2=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\ldots}}}} = \sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^{\sqrt[4]{4}^\ldots}}}} = 4$