How to solve $x^{x^x}=(x^x)^x$? How can we solve the equation : 
$x^{x^x}=(x^x)^x$
with $x \in {\mathbb{R}+}^*$
Thanks for heping me :)
 A: There is no $x\in \mathbb R^+$ aside from $1$ or $2$ that satisfies the equation. It is fairly easy to see this. Assume that $x$ is not $1$ or $2$. Then, $x$ is non-zero.
Write $$x^{x^x}=(x^x)^x$$
As $$x^{x^x}=x^{x^2}$$
Since $x$ is non-zero, we may divide both side by $x^{x^2}$:
$$x^{x^x-x^2}=1$$
Since $x\in \mathbb R^+$, we must then have $x^x-x^2=0\implies x^x=x^2$- the only other possibility is $x=1$ which we have assumed not to be true.
We may again use the fact that $x$ is non-zero to divide both sides by $x^2$. This gives us $$x^{x-2}=1$$
Again, we're working in $\mathbb R^+$, so this must mean $x-2=0$ (we assumed $x\neq 1$), i.e. $x=2$ but this contradicts our initial assumption.
A: We can check when the exponents are equal.
It is $x^{x^x}=(x^x)^x\Leftrightarrow x^{(x^x)}=x^{x^2}\Leftrightarrow x^x=x^2$
Now x^x-x^2=0\Leftrightarrow $x^2(x^{x-2}-1)=0$.
So $x^2=0$ or $x^{x-2}=1$.
Since $x\neq 0$ we have $x^{x-2}=1$ left, which holds if $x-2=0$. So $x=2$
And $2^{2^2}=2^{2\cdot 2}$
Edit: And x=1 is an obvious solution...
A: This is the same as
$$x^{x^x}=x^{x^2}$$
$$x^x=x^2$$
$$x^{x-2}=1$$
$$\therefore x=1,2$$
A: Applying $\log$ to both sides
$$
x^x\log x = x^2\log x\Rightarrow (x^x-x^2)\log x=0
$$
so we have $\log x= 0\Rightarrow x = 1$ and $x^x=x^2$ with two solutions $x = 1$ and $x = 2$ because following $x\log x = 2\log x\Rightarrow (x-2)\log x = 0$
A: You can do this by noting $(x^x)^x=x^{2x}$, and then by comparing the exponents, since the exponential with base x is increasing.
(Note, writing ${x^x}^x$ is extremely ambiguous. Either you mean $(x^x)^x$ or $x^{(x^x)}$. In this case the context is clear, but in general it is safer to use brackets.)
