Solving $x^{\sqrt{x}}=\sqrt{x^x}$ Considering ${\sqrt{x}}=y$, what is the resolution process of $x^{\sqrt{x}}=\sqrt{x^x}$, with results S={1,4}?
 A: Square both sides to get
$$x^{2\sqrt x}=x^x$$
Either $x=1$, or $2\sqrt x=x\implies x=4$.
A: Take logs of both sides to get:
$\sqrt x \log x = \frac 12 x\log x$
which has solutions of either $\log x = 0 \implies x = 1$
or:
$\sqrt x = \frac 12 x$
which has solutions of $x = 0,4$. Exclude $x=0$ as it is inadmissible in the original equation, leaving the solutions $x = 1,4$.
A: Assuming we are only consider real numbers here.
$$x^{\sqrt{x}}=\sqrt{x^x}$$
Firstly, notice that $x\ge 0$, in order for $\sqrt{x}$ to make sense. Also $0^0$ is not defined, thus $x\ne 0$, that is, $x >0$.
Now square both sides:
$$x^{2\sqrt{x}}=x^x$$
Take $\log$ on both sides:
$$2\sqrt{x}\log x=x \log x$$
Thus we have:
$$4 x = x^2$$ 
Or
$$\log x = 0$$
So $x = 1$ or $x=4$
A: Its natural to assume $x> 0$. So let's square both sides to get
$$x^{2\sqrt x}=x^x.$$
Its clear, that $x^x>0$, so use the logarithm to obtain
$$ 2\sqrt x \cdot\log(x)=x \cdot \log(x).$$
Here, you have to exclude $x=1$. Simple substitution of $x=1$ reveals that this would already be a solution. Divide by $\log(x)$. It remains
$$2\sqrt x=x$$
which only has the solution $x=4$ for positive $x$. 
