Question

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


​

Answer 1

Explanation:

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

Answer 2

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$\huge\star\pink{solution}$

$= > x {}^{ \sqrt{x} } = \sqrt{x} {}^{x} \\ \\ = > x {}^{ \sqrt{x} } = x {}^{ \frac{x}{2} } \\ \\ \: rhs \: and \: lhs \: base \: are \: same \\ so \: power \: rule \: compairison \\ rhs \: and \: lhs \: power \\ \\ = > \sqrt{x} = \frac{x}{2} \\ \\ = > x = 2 \sqrt{x}$

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uses

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

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I hope it helps you