Solving equation with infinite exponent tower How to solve this equation for $x$ where $a&gt;0$? 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?
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:
$$ln(a^\frac{1}{a}) = ln(x)$$
$$a^\frac{1}{a} = x$$
$$\sqrt [a]{a} = x$$
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.