Algebra exercise Level 1

Exercise:

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

Solution: Take logarithm both sides to get $\sqrt{x}\ln x=x\ln \sqrt{x}$

$\Leftrightarrow \sqrt{x}\ln x-x\ln \sqrt{x}=0$ $\Leftrightarrow \sqrt{x}\ln x-\frac{x}{2}\ln x=0$ $\ln x\left( \sqrt{x}-\frac{x}{2} \right)=0$

So $\ln x=0\,\,\,or\,\sqrt{x}-\frac{x}{2}=0\Leftrightarrow \ln x=\ln 1\,\,or\,\,\sqrt{x}=\frac{x}{2}$

 $\Leftrightarrow x=1\,\,or\,\,x=\frac{{{x}^{2}}}{4}\Leftrightarrow x=1\,\,\,or\,\,4x={{x}^{2}}$

Thus $4x-{{x}^{2}}=0\Leftrightarrow x\left( 4-x \right)=0\Leftrightarrow x=0\,\,or\,\,x=4$

 but $x=0$ is rejected so the roots are $x=1\,\,or\,\,4$