Question Number 195002 by mathlove last updated on 22/Jul/23
$${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$
Answered by dimentri last updated on 22/Jul/23
$$\:\:\:\left({x}\right)^{\sqrt{{x}}} \:=\:\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}{x}} \\ $$$$\:\:\left(\mathrm{1}\right)\:\sqrt{{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\Rightarrow{x}−\mathrm{2}\sqrt{{x}}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\sqrt{{x}}\:\left(\sqrt{{x}}−\mathrm{2}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{4} \\ $$$$\:\:\:\left(\mathrm{2}\right)\:{x}=\:\mathrm{1} \\ $$$$\:\:\therefore\:{x}\:\in\:\left\{\mathrm{1},\:\mathrm{4}\right\}\:\: \\ $$$$\:\: \\ $$
Commented by mathlove last updated on 22/Jul/23
$${thanks} \\ $$