Issues solving equations involving $x^{x^x...}$?

Question

I stumbled across this problem:

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

Obviously, I used the substitution trick and I got

$x^2=2$

and thus, $x=\pm\sqrt{2}$. I have tested that this works.

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However, I tried to solve

$x^{x^{x^{...}}}=4$

and it yields the same real answers ($x^4=4$). I have no idea why this is.

Answer 1

The infinite tetration $f(x)=x^{x^{x^\cdots}}$ only converges for $(1/e)^e \leq x \leq e^{e-1}$ and assumes values in $[1/e,e]$. Thus the inverse function $f^{-1}(y)$ is only defined for $y\in [1/e,e]$. So the solution of the first example is $\sqrt{2}$ whereas the second example does not have a solution (since $4>e$).

Answer 2

Note that $x^{x^{x^{\vdots}}} = 4$ implies that $x^4 = 4$, but not the other way around.

So any solution to the first equation must be a solution to the second equation. However, not all solutions to the second equation will be solutions to the first equation, i.e. the second equation may have some extraneous solutions.

Note: This is similar to what is going on in this other question.