Question Number 17638 by chux last updated on 08/Jul/17
$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{this} \\ $$$$\mathrm{confusing}\:\mathrm{question} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2x}/\mathrm{y}} ×\mathrm{y}^{\mathrm{y}/\mathrm{x}} =\mathrm{4}……\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\left(\mathrm{xy}\right)^{\mathrm{xy}+\mathrm{yx}} =\mathrm{16}…..\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$
Commented by mrW1 last updated on 09/Jul/17
$$\mathrm{x}=\mathrm{1}\:\mathrm{and}\:\mathrm{y}=\mathrm{2} \\ $$
Commented by chux last updated on 09/Jul/17
$$\mathrm{please}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{analytically}. \\ $$$$ \\ $$
Commented by mrW1 last updated on 09/Jul/17
![from (2) you get [(xy)^((xy)) ]^2 =16 (xy)^((xy)) =4 (xy)=2 from (1) you get x^(2x/y) =4y^(−y/x) (x^(2x/y) )^(xy) =(4y^(−y/x) )^(xy) x^(2x^2 ) =16y^(−y^2 ) x^(2x^2 ) =16((x/2))^(2/x^2 ) x^(2(x^2 −(1/x^2 ))) =2^(2(1−(1/x^2 ))) ⇒x=1 and 1−(1/x^2 )=0](https://www.tinkutara.com/question/Q17667.png)
$$\mathrm{from}\:\left(\mathrm{2}\right)\:\mathrm{you}\:\mathrm{get} \\ $$$$\left[\left(\mathrm{xy}\right)^{\left(\mathrm{xy}\right)} \right]^{\mathrm{2}} =\mathrm{16} \\ $$$$\left(\mathrm{xy}\right)^{\left(\mathrm{xy}\right)} =\mathrm{4} \\ $$$$\left(\mathrm{xy}\right)=\mathrm{2} \\ $$$$\mathrm{from}\:\left(\mathrm{1}\right)\:\mathrm{you}\:\mathrm{get} \\ $$$$\mathrm{x}^{\mathrm{2x}/\mathrm{y}} =\mathrm{4y}^{−\mathrm{y}/\mathrm{x}} \\ $$$$\left(\mathrm{x}^{\mathrm{2x}/\mathrm{y}} \right)^{\mathrm{xy}} =\left(\mathrm{4y}^{−\mathrm{y}/\mathrm{x}} \right)^{\mathrm{xy}} \\ $$$$\mathrm{x}^{\mathrm{2x}^{\mathrm{2}} } =\mathrm{16y}^{−\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{x}^{\mathrm{2x}^{\mathrm{2}} } =\mathrm{16}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{x}^{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)} =\mathrm{2}^{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{1}\:\mathrm{and}\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{0} \\ $$
Commented by chux last updated on 09/Jul/17
$$\mathrm{wow}!!\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}. \\ $$