Find x in infinite sequence

Question

We have: $$ x^{x^{x^{ x^{x ^{x ^{\dots}}}}}} = 2.$$

I tried a reasoning by recursion:

For $n=1$: \begin{align} x^x &= 2 \\ \implies x\ln x &= \ln 2 \end{align}

For $n=2$: \begin{align} x^{x^x} &= 2 \\ \implies x\ln x^x &= \ln 2 \\ \implies x^2\ln x &= \ln 2 \\ \end{align}

For an arbitrary $n$ we can solve: $$x^n \ln x = \ln 2 $$

But for $n \to \infty$, I can't seem to find something. Does anyone have an idea on how to tackle this problem?

Answer 1

Hint: If the solution exists then

$$x^{x^{x^{\cdots}}}=x^{(x^{x^{\cdots}})}=x^{(2)}=x^2=2.$$

The stuff in the power is equal to $2$ hence we can replace it by $2$.