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x-y-y-x-x-2-y-3-find-x-and-y-




Question Number 11805 by tawa last updated on 01/Apr/17
x^y  = y^x       x^2  = y^3   find x and y
$$\mathrm{x}^{\mathrm{y}} \:=\:\mathrm{y}^{\mathrm{x}} \:\:\:\: \\ $$$$\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$
Answered by mrW1 last updated on 01/Apr/17
(i)⇒x=y^(x/y)   (ii)⇒x=y^(3/2)   ⇒(x/y)=(3/2)  ⇒x=(3/2)y  (3/2)y=y^(3/2)   (3/2)=y^(1/2)   ⇒y=((3/2))^2 =(9/4)  ⇒x=(3/2)×(9/4)=((27)/8)
$$\left({i}\right)\Rightarrow{x}={y}^{\frac{{x}}{{y}}} \\ $$$$\left({ii}\right)\Rightarrow{x}={y}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}{y} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{y}={y}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}={y}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{4}}=\frac{\mathrm{27}}{\mathrm{8}} \\ $$
Commented by tawa last updated on 01/Apr/17
wow, God bless you sir.
$$\mathrm{wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Apr/17
x=y=1 also is an answer.
$${x}={y}=\mathrm{1}\:{also}\:{is}\:{an}\:{answer}. \\ $$
Commented by mrW1 last updated on 01/Apr/17
that′s right.
$${that}'{s}\:{right}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Apr/17
x^(2y) =y^(2x) ⇒(x^2 )^y =y^(2x) ⇒(y^3 )^y =y^(2x)   3y=2x⇒((3/2)y)^2 =y^3 ⇒9y^2 =4y^3 ⇒  y^2 (4y−9)=0⇒y=(9/4)⇒x=(3/2)×(9/4)=((27)/8)
$${x}^{\mathrm{2}{y}} ={y}^{\mathrm{2}{x}} \Rightarrow\left({x}^{\mathrm{2}} \right)^{{y}} ={y}^{\mathrm{2}{x}} \Rightarrow\left({y}^{\mathrm{3}} \right)^{{y}} ={y}^{\mathrm{2}{x}} \\ $$$$\mathrm{3}{y}=\mathrm{2}{x}\Rightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}{y}\right)^{\mathrm{2}} ={y}^{\mathrm{3}} \Rightarrow\mathrm{9}{y}^{\mathrm{2}} =\mathrm{4}{y}^{\mathrm{3}} \Rightarrow \\ $$$${y}^{\mathrm{2}} \left(\mathrm{4}{y}−\mathrm{9}\right)=\mathrm{0}\Rightarrow{y}=\frac{\mathrm{9}}{\mathrm{4}}\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{4}}=\frac{\mathrm{27}}{\mathrm{8}} \\ $$

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