Question

If $(2x{)}^{{\log }_{\sqrt{x}}x}=16$, then find the value of x.

Answer 1

$(2x{)}^{{\log }_{\sqrt{x}}x}=16$

$(2x{)}^{2{\log }_{x}x}=16$

$(2x{)}^2=16$

$4x^2=16$

$x^2=4$

$x=\pm 2$

When $x=-2,{\log }_{\sqrt{x}}x$ is not defined.

Therefore, $x=2$.