Question

find
${x}^{ \sqrt{x} } = \sqrt{ {x}^{x} }$
​

Answer 1

ㅤㅤㅤㅤANSWER

${x}^{ \sqrt{x} } = \sqrt{ {x}^{x} }$

$= > {x}^{ {x}^{ \frac{1}{2} } } = {{(x}^{x} )}^{ \frac{1}{2} }$

$= > {x}^{ {x}^{ \frac{1}{2 } } } = {x}^{ \frac{x}{2} }$

Bases are same

$= > {x}^{ \frac{1}{2} } = \frac{x}{2}$

Square both sides

$= > {( {x}^{ \frac{1}{2} } )}^{2} = ( { \frac{x}{2} })^{2}$

$= > x = \frac{ {x}^{2} }{4}$

$= > 4x = {x}^{2}$

$= > 4x = x \times x$

$= > \frac{4x}{x} = x$

$= > x = 4$

Ans. x=4