Simple Algebra exercise asked in the brilliant website

Exercise:

Solve in $\mathbb{R}$ ,${{x}^{\sqrt{x}}}={{\left( \sqrt{x} \right)}^{x}}$

Solution : we have ${{x}^{\sqrt{x}}}={{\left( \sqrt{x} \right)}^{x}}$ taken logarithm both sides to get $\sqrt{x}\ln x=x\ln \sqrt{x}=\frac{x}{2}\ln x$

So $\left( 2\sqrt{x}-x \right)\ln x=0$ $\Leftrightarrow 2\sqrt{x}-x=0\,\,\,or\,\,\,\ln x=0$ $\Leftrightarrow 4x-{{x}^{2}}=0\,\,\,or\,\,\,\ln x=\ln 1$

$\Rightarrow x=1\,\,or\,\,x\left( 4-x \right)=0$ $\Leftrightarrow x=0\,\,\,or\,\,\,x=4$ but $x\ne 0$ so the accepted roots are $x=\left\{ 1,4 \right\}$