In $x^{x^{x^{x^{...}}}}=2$, the answer is $x=\sqrt{2}$. Why not also $-\sqrt{2}$?

Question

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$$x^{x^{x^{x^{...}}}}=2$$

The answer is clearly $\sqrt{2}$, but I'm curious why it's not also $-\sqrt{2}$.

Am I missing something basic?

Answer 1

You have $x^{x^{x^{\dots}}} = 2$. You can simply substitute in the exponent to conclude that $x^2 = 2$. But that's only one way. If, instead you have that $x^2=2$, it's not necessarily true that $x^{x^{x^{\dots}}} = 2$. So solving $x^2=2$ lets us find candidate solutions for the original equation, but then we need to confirm that each candidate is in fact a solution.

First, let's check whether $x=\sqrt{2}$ satisfy $x^{x^{x^{\dots}}} = 2$. For this we need to know what $x^{x^{x^{\dots}}} = 2$ actually is. We can define it as the limit of the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, \dots$, and we can check that, if $x=\sqrt{2}$, this sequence indeed converges to 2.

What about $x=-\sqrt{2}$? Consider $(-\sqrt{2})^{-\sqrt{2}}$. This is not even defined in real numbers. See How do you compute negative numbers to fractional powers? So we need to go to complex numbers. Then $(-\sqrt{2})^{-\sqrt{2}}$ actually takes infinitely many values, which causes problems on its own. EDIT: We can consider the sequence of principal values and try to check if it converges, but that turns out to be more complex than I thought.

Answer 2

The tower is $\lim_{n\to\infty}u_n$ with $u_1=x,\,u_{n+1}=x^{u_n}$. If $x=\sqrt{2}$ this is an increasing sequence of positive numbers $<2$. If $x=-\sqrt{2}$, on the other hand, it quickly goes off the rails. What is $(-\sqrt{2})^{-\sqrt{2}}$ supposed to be?