Question Number 7222 by Tawakalitu. last updated on 16/Aug/16
$${What}\:{are}\:{the}\:{possible}\:{solution}\:{satisfying} \\ $$$${x}^{{y}} \:=\:{y}^{{x}} \\ $$
Commented by Tawakalitu. last updated on 17/Aug/16
$${I}\:{appreciate}\:{your}\:{effort}.\:{thanks}\:{sir}. \\ $$
Commented by Yozzia last updated on 17/Aug/16
$${x}={y}\in\mathbb{C}\:{constitutes}\:{part}\:{of}\:{the}\:{solution} \\ $$$${set}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${x}^{{y}} ={y}^{{x}} \\ $$$${ylnx}={xlny} \\ $$$$−\frac{\mathrm{1}}{{x}}{lnx}=−\frac{\mathrm{1}}{{y}}{lny} \\ $$$${x}^{−\mathrm{1}} {lnx}^{−\mathrm{1}} ={y}^{−\mathrm{1}} {lny}^{−\mathrm{1}} \\ $$$${e}^{{lnx}^{−\mathrm{1}} } {lnx}^{−\mathrm{1}} ={e}^{{lny}^{−\mathrm{1}} } {lny}^{−\mathrm{1}} \\ $$$${W}\left({e}^{{lnx}^{−\mathrm{1}} } {lnx}^{−\mathrm{1}} \right)={W}\left({e}^{{lny}^{−\mathrm{1}} } {lny}^{−\mathrm{1}} \right) \\ $$$$\Rightarrow{lnx}^{−\mathrm{1}} ={lny}^{−\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{y}}\Rightarrow{x}={y}\in\mathbb{C}. \\ $$$${By}\:{this}\:{working},\:{x}^{{y}} ={y}^{{x}} \:\Leftrightarrow{x}={y}. \\ $$