Question

The roots of the equation ........are
${x}^{ \sqrt{x} } = { \sqrt{x} }^{x}$



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Answer 1

Step-by-step explanation:

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

Put logarithm on both sides we get

√x log(x) = x/2 log(x) => log(x) (√x -x/2) =0 So if log(x) = 0 => x=1

(√x - x/2)=0=> √x = x/2 => 2 = x/√x =√x =>

Square on both sides we get x=4. Hence the solutions to the equation

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

are 1 and 4 respectively. Hope it helps you.