TITLE: Find the number of roots of the equation in $\mathbb{R}$
QUESTION [5 upvotes]: How many roots does the equation $$\\x^{x^x}=(x^x)^x\\$$ have in $\\\mathbb{R}$?
My observations:I observed that $x=-1,1,2$ are its roots.
Are there other roots of this equation?And how we can find them?
Thanks.

REPLY [7 votes]: $$x^{x^x}=(x^x)^x$$
$$x^{x^x}=x^{x^2}$$
Taking logarithm on both sides,
$${x^x}\,log\, \left\lvert x \right\rvert={x^2}\,log\, \left\lvert x \right\rvert$$
$$({x^x}-{x^2})\,log\,\left\lvert x \right\rvert=0$$
Considering $log\, \left\lvert x \right\rvert=0$, we get
$$x=\pm\,1$$
Considering $${x^x}-{x^2}=0$$
$${x^x}={x^2}$$
Taking logarithm on both sides,
$$x\,log\,\left\lvert x \right\rvert=2\,log\,\left\lvert x \right\rvert$$
$$(x-2)\,log\,\left\lvert x \right\rvert=0$$
Therefore,
$$x\,=-1\,,\,1,\,2$$