$x=2^{x^{x^{.^{.^{.}}}}}$ My attempt at solving the question: $$\log_2x=x^{x^{x^{.^{.^{.}}}}}$$ Now here I use a trivial trick which I have seen in my class to substitute infinite powers. $$\log_2x=x^{\log_2x}$$ After that I have no idea how to solve this equation.
I opened the graph and solution of the question on Wolfram|Alpha but I could not understand what is written.
On
Your right side only makes sense as a limit of a tower, the limit of $$2, 2^x, 2^{x^x}, 2^{x^{x^x}} \ldots$$ If there is such a limit $L$, we must have $L^x=L$, which can be satisfied for $L=0,1$ or for $x=1$. You should be able to convince yourself that all the terms of the series are greater than $1$, but if $x \lt 1$ the series will converge to $1$. Unfortunately, if $x=1$ the series is a constant $2$. In neither case does the tower converge to $x$, so the equation has no real solution.
The equation you wrote is equivalent to $$x=2^{x^\infty}, $$ which has no solution. (If $x<1$, it reduces to $x=2^0=1$, if $x=1$ to $x=2^1=2$ and $x>1$ to $x=2^\infty=\infty$.)