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Question Number 6922 by Tawakalitu. last updated on 02/Aug/16

x^y  = y^x     make x the subject of the formular

$${x}^{{y}} \:=\:{y}^{{x}} \\ $$$$ \\ $$$${make}\:{x}\:{the}\:{subject}\:{of}\:{the}\:{formular} \\ $$

Commented by Yozzii last updated on 03/Aug/16

x^y =y^x   ylnx=xlny  ((lnx)/x)=((lny)/y)  Let u=lnx⇒x=e^u ⇒x^(−1) =e^(−u)   ∴ ue^(−u) =((lny)/y)  −ue^(−u) =−((lny)/y)  ∴W(−ue^(−u) )=W(((−lny)/y))  −u=W(ln(1/y^(1/y) ))  u=−W(−lny^(1/y) )  lnx=−W(−lny^(1/y) )  ⇒x=e^(−W(−lny^(1/y) ))

$${x}^{{y}} ={y}^{{x}} \\ $$$${ylnx}={xlny} \\ $$$$\frac{{lnx}}{{x}}=\frac{{lny}}{{y}} \\ $$$${Let}\:{u}={lnx}\Rightarrow{x}={e}^{{u}} \Rightarrow{x}^{−\mathrm{1}} ={e}^{−{u}} \\ $$$$\therefore\:{ue}^{−{u}} =\frac{{lny}}{{y}} \\ $$$$−{ue}^{−{u}} =−\frac{{lny}}{{y}} \\ $$$$\therefore{W}\left(−{ue}^{−{u}} \right)={W}\left(\frac{−{lny}}{{y}}\right) \\ $$$$−{u}={W}\left({ln}\frac{\mathrm{1}}{{y}^{\mathrm{1}/{y}} }\right) \\ $$$${u}=−{W}\left(−{lny}^{\mathrm{1}/{y}} \right) \\ $$$${lnx}=−{W}\left(−{lny}^{\mathrm{1}/{y}} \right) \\ $$$$\Rightarrow{x}={e}^{−{W}\left(−{lny}^{\mathrm{1}/{y}} \right)} \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 03/Aug/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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