Find the number of roots of the equation in $\mathbb{R}$ 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.
 A: $$x^{x^x}=(x^x)^x$$
$$x^{x^x}=x^{x^2}$$
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,
$$x\,=-1\,,\,1,\,2$$
A: You have them all.  If $x \lt 0$, $x^x$ is only defined for $x$ integral.  If $x$ is integral and less than $-1$, $x^x$ is not integral and $x^{x^x}$ is not defined.  Similarly neither side is defined for $x=0$. The only solution less than or equal to zero is $-1$.  
For $x \gt 0$, we can write this as $x^{(x^x)}=x^{(x^2)}$  and take natural logs to give $x^x \ln x = x^2 \ln x$.  Since $\ln x \neq 0$ unless $x=1$, we can check that case, finding it is a solution, then exclude it and divide by it. This gives $x^x=x^2, x \gt 0 ,x \neq 1$  Taking another log gives $x=2$ as the only remaining solution.
A: For $ x^{x^x} = x^{x^2} $ the three roots are:  1 ( double root), and 2, found by substitution possible Taylor expansion around root and direct graphing. 
