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Question Number 17576 by chux last updated on 07/Jul/17

$$\mathrm{solve}\mathrm{the}\mathrm{equation}$$$$$$$${\mathrm{x}}^{\mathrm{y}}={\mathrm{y}}^{\mathrm{x}}.......\left(1\right)$$$$$$$${\mathrm{x}}^2={\mathrm{y}}^3.......\left(2\right)$$

Answered by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17

$${\left(x^2\right)}^y={\left(y^3\right)}^y\Rightarrow {\left(x^y\right)}^2=y^{3y}\Rightarrow {\left(y^x\right)}^2=y^{3y}$$$$\Rightarrow y^{2x}=y^{3y}\Rightarrow 2x=3y$$$$\Rightarrow {\left(\frac{3}{2}y\right)}^2=y^3\Rightarrow y^2(y-\frac{9}{4})=0\Rightarrow y=0,\frac{9}{4}$$$$\Rightarrow x=0,\frac{3}{2}.\frac{9}{4}=0,\frac{27}{8}.$$$$\Rightarrow (x,y)=(\frac{27}{8},\frac{9}{4}).$$$$note:(x,y)=(1,1),isalsoasolution.$$$$but:(x,y)=(0,0),notacceptable.$$

Commented by chux last updated on 08/Jul/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by chux last updated on 08/Jul/17

$$\mathrm{please}\mathrm{can}\mathrm{it}\mathrm{be}\mathrm{solved}\mathrm{analytically}$$$$\mathrm{to}\mathrm{get}(\mathrm{x},\mathrm{y})=(1,1)$$

Commented by mrW1 last updated on 08/Jul/17

$$\mathrm{from}{y}^{2x}=y^{3y}$$$$\Rightarrow 2x=3y...\left(\mathrm{i}\right)$$$$\mathrm{or}$$$$\Rightarrow \mathrm{y}=1...\left(\mathrm{ii}\right)$$$$$$$$\mathrm{from}\left(\mathrm{ii}\right)$$$$\Rightarrow \mathrm{y}=1$$$$\Rightarrow \mathrm{x}=1$$

Commented by chux last updated on 08/Jul/17

$$\mathrm{ok}....\mathrm{but}\mathrm{can}\mathrm{lambert}\mathrm{W}\mathrm{function}$$$$\mathrm{solve}\mathrm{it}?$$

Commented by mrW1 last updated on 09/Jul/17

$$\mathrm{you}\mathrm{have}\mathrm{seen}\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{without}$$$$\mathrm{lambert}\mathrm{function}.$$

Commented by chux last updated on 09/Jul/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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