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Question-190397




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   ⇒y=x^(y/x)   ((x/y))^(x/y)   =((x/x^(y/x) ))^(x/y)   =(x^(1−(y/x)) )^(x/y)   =x^(((x−y)/x)×(x/y)) =x^((x−y)/y)   given y=2x   ((x/y))^(x/y) =x^((x−y)/y)   ⇒((x/(2x)))^(x/(2x)) =(x)^((x−2x)/(2x))   ⇒((1/2))^(1/2) =(x)^((−1)/2)   ⇒(1/( (√2)))=(1/( (√x)))  ⇒x=2∴y=2x=2×2=4
$$\:{x}^{{y}} ={y}^{{x}} \\ $$$$\Rightarrow{y}={x}^{\frac{{y}}{{x}}} \\ $$$$\left(\frac{{x}}{{y}}\right)^{\frac{{x}}{{y}}} \\ $$$$=\left(\frac{{x}}{{x}^{\frac{{y}}{{x}}} }\right)^{\frac{{x}}{{y}}} \\ $$$$=\left({x}^{\mathrm{1}−\frac{{y}}{{x}}} \right)^{\frac{{x}}{{y}}} \\ $$$$={x}^{\frac{{x}−{y}}{{x}}×\frac{{x}}{{y}}} ={x}^{\frac{{x}−{y}}{{y}}} \\ $$$${given}\:{y}=\mathrm{2}{x} \\ $$$$\:\left(\frac{{x}}{{y}}\right)^{\frac{{x}}{{y}}} ={x}^{\frac{{x}−{y}}{{y}}} \\ $$$$\Rightarrow\left(\frac{{x}}{\mathrm{2}{x}}\right)^{\frac{{x}}{\mathrm{2}{x}}} =\left({x}\right)^{\frac{{x}−\mathrm{2}{x}}{\mathrm{2}{x}}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left({x}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$$\Rightarrow{x}=\mathrm{2}\therefore{y}=\mathrm{2}{x}=\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$

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