Find $x$ in $\large \,\, A^{A^{{{A^.}^.}^.}}= \,\,(\sqrt [x\cdot x\cdot x\cdot ...] x)^{ (((1\cdot x+1)x +1)x +1)x+1...} $

Question

If $\large \,\, A^{A^{{{A^.}^.}^.}}= \,\,(\sqrt [x\cdot x\cdot x\cdot ...] x)^{ (((1\cdot x+1)x +1)x +1)x+1...} $, and $\large \,\,A = (\sqrt[3]{3\sqrt 3 })^{\frac{\sqrt 3}{3}} $, find $x$

I have no idea how to start it.

But my book says that $x=3$

Thank you very much.

Answer 1

We have $A=(\sqrt3)^\tfrac1{\sqrt3}$ , which, when infinitely tetrated, converges to $A^{A^{A^{\cdot^{\cdot^\cdot}}}}=\sqrt3$. On the other hand, the exponent of x is $\displaystyle\lim_{n\to\infty}\frac{1+x+x^2+\ldots+x^{n-1}}{x^n}=\lim_{n\to\infty}\frac{x^n-1}{(x-1)x^n}=\lim_{n\to\infty}\frac{1-x^{-n}}{x-1}=$ $=\dfrac1{x-1}$ when $|x|>1$, and divergent otherwise. So $x^{^\tfrac1{x-1}}=\sqrt3=3^{^\tfrac12}=3^{^\tfrac1{3-1}}$. I trust that you can take it from here... :-)