Question Number 190397 by pascal889 last updated on 02/Apr/23
Answered by som(math1967) last updated on 02/Apr/23
$$\:{x}^{{y}} ={y}^{{x}} \\ $$$$\Rightarrow{y}={x}^{\frac{{y}}{{x}}} \\ $$$$\left(\frac{{x}}{{y}}\right)^{\frac{{x}}{{y}}} \\ $$$$=\left(\frac{{x}}{{x}^{\frac{{y}}{{x}}} }\right)^{\frac{{x}}{{y}}} \\ $$$$=\left({x}^{\mathrm{1}−\frac{{y}}{{x}}} \right)^{\frac{{x}}{{y}}} \\ $$$$={x}^{\frac{{x}−{y}}{{x}}×\frac{{x}}{{y}}} ={x}^{\frac{{x}−{y}}{{y}}} \\ $$$${given}\:{y}=\mathrm{2}{x} \\ $$$$\:\left(\frac{{x}}{{y}}\right)^{\frac{{x}}{{y}}} ={x}^{\frac{{x}−{y}}{{y}}} \\ $$$$\Rightarrow\left(\frac{{x}}{\mathrm{2}{x}}\right)^{\frac{{x}}{\mathrm{2}{x}}} =\left({x}\right)^{\frac{{x}−\mathrm{2}{x}}{\mathrm{2}{x}}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left({x}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$$\Rightarrow{x}=\mathrm{2}\therefore{y}=\mathrm{2}{x}=\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$