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Question Number 190397 by pascal889 last updated on 02/Apr/23

Answered by som(math1967) last updated on 02/Apr/23

$$x^y=y^x$$$$\Rightarrow y=x^{\frac{y}{x}}$$$${\left(\frac{x}{y}\right)}^{\frac{x}{y}}$$$$={\left(\frac{x}{x^{\frac{y}{x}}}\right)}^{\frac{x}{y}}$$$$={\left(x^{1-\frac{y}{x}}\right)}^{\frac{x}{y}}$$$$=x^{\frac{x-y}{x}\times \frac{x}{y}}=x^{\frac{x-y}{y}}$$$$giveny=2x$$$${\left(\frac{x}{y}\right)}^{\frac{x}{y}}=x^{\frac{x-y}{y}}$$$$\Rightarrow {\left(\frac{x}{2x}\right)}^{\frac{x}{2x}}={\left(x\right)}^{\frac{x-2x}{2x}}$$$$\Rightarrow {\left(\frac{1}{2}\right)}^{\frac{1}{2}}={\left(x\right)}^{\frac{-1}{2}}$$$$\Rightarrow \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{x}}$$$$\Rightarrow x=2\therefore y=2x=2\times 2=4$$

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