Will we consider -√2 as a solution?

Question

For the equation

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

we will get two solutions $1$ and $\sqrt 2$ by this method

$${x^{x^{x^{\mathstrut^{.^{.^{.}}}}}}} =x^2 \iff x^{x^2} = x^2 \iff x^2 \ln x=2 \ln x$$

now either $\ln(x)=0$ which will give solution $x= 1$

or $x^2=2$ which will give solution $\sqrt 2$ and $-\sqrt2$

so will we consider $-\sqrt 2$ as its solution or not?

Edit: exponent extands to infinity rather being finite

Answer 1

For $x\in \mathbb{R}$ we have that:

$x^{x^2} = x^2 \iff x^2 \ln x=2 \ln x \quad$ iff $x>0$

Answer 2

As a note, I'd first study the real-valued function $f:{\Bbb R}_{\geq 0}\rightarrow {\Bbb R}:x\mapsto x^x$ which can be done by first semester calculus.

Next you could extend to the full domain $\Bbb R$ which gives rise to complex function values.

Answer 3

You need to either state how many "x"s there are or extend the exponent stack to infinity: $x^{x^{x^{x^{\cdot^{\cdot^{\cdot^{x^{\cdot^{\cdot^{\cdot}}}}}}}}}}$