Question Number 171927 by Mikenice last updated on 21/Jun/22
$${solve}: \\ $$$${x}^{\sqrt{{x}}} =\left(\sqrt[{\mathrm{4}}]{{x}}\right)^{\mathrm{1}+\sqrt{{x}}} \\ $$
Answered by aleks041103 last updated on 22/Jun/22
$${x}^{\sqrt{{x}}} ={x}^{\frac{\mathrm{1}+\sqrt{{x}}}{\mathrm{4}}} \\ $$$$\Rightarrow\sqrt{{x}}=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{{x}}}{\mathrm{4}} \\ $$$$\Rightarrow\sqrt{{x}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$
Commented by mr W last updated on 22/Jun/22
$${x}=\mathrm{1}\:{as}\:{well} \\ $$