# Find x in infinite sequence [duplicate]

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?

## marked as duplicate by Martin R, Community♦Apr 18 at 7:38

$$x^{x^{x^{\cdots}}}=x^{(x^{x^{\cdots}})}=x^{(2)}=x^2=2.$$
The stuff in the power is equal to $$2$$ hence we can replace it by $$2$$.