Question

Please solve this question..
x
${x}^{x \sqrt{x} } = {x \sqrt{x} }^{x}$

Answer 1

Answer : (D) 64/27

________________________

Solution :

________________

${x}^{x \sqrt[3]{x} } = ({x. \sqrt[3]{x}) }^{x} \\ \\ = > {x}^{x. {x}^{ \frac{1}{3} } } = {(x. {x}^{ \frac{1}{3} } )}^{x} \\ \\ = > {x}^{ {x}^{ \frac{4}{3} } } = {x}^{ \frac{4x}{3} } \\ \\ = > {x}^{ \frac{4}{3} } = \frac{4x}{3}$

Taking cube both the sides :

${x}^{4} = \frac{64 {x}^{3} }{27} \\ \\ = > 27 {x}^{4} = 64 {x}^{3} \\ \\ = > 27x = 64 \\ \\ = > x = \frac{64}{27}$