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Question Number 17638 by chux last updated on 08/Jul/17

$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{this}$$$$\mathrm{confusing}\mathrm{question}$$$$$$$${\mathrm{x}}^{2\mathrm{x}/\mathrm{y}}\times {\mathrm{y}}^{\mathrm{y}/\mathrm{x}}=4......\left(1\right)$$$$$$$${\left(\mathrm{xy}\right)}^{\mathrm{xy}+\mathrm{yx}}=16.....\left(2\right)$$$$$$$$\mathrm{solve}\mathrm{for}\mathrm{x}\mathrm{and}\mathrm{y}$$

Commented by mrW1 last updated on 09/Jul/17

$$\mathrm{x}=1\mathrm{and}\mathrm{y}=2$$

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

$$\mathrm{from}\left(2\right)\mathrm{you}\mathrm{get}$$$${\left[{\left(\mathrm{xy}\right)}^{\left(\mathrm{xy}\right)}\right]}^2=16$$$${\left(\mathrm{xy}\right)}^{\left(\mathrm{xy}\right)}=4$$$$\left(\mathrm{xy}\right)=2$$$$\mathrm{from}\left(1\right)\mathrm{you}\mathrm{get}$$$${\mathrm{x}}^{2\mathrm{x}/\mathrm{y}}={4\mathrm{y}}^{-\mathrm{y}/\mathrm{x}}$$$${\left({\mathrm{x}}^{2\mathrm{x}/\mathrm{y}}\right)}^{\mathrm{xy}}={\left({4\mathrm{y}}^{-\mathrm{y}/\mathrm{x}}\right)}^{\mathrm{xy}}$$$${\mathrm{x}}^{{2\mathrm{x}}^2}={16\mathrm{y}}^{-{\mathrm{y}}^2}$$$${\mathrm{x}}^{{2\mathrm{x}}^2}=16{\left(\frac{\mathrm{x}}{2}\right)}^{\frac{2}{{\mathrm{x}}^2}}$$$${\mathrm{x}}^{2({\mathrm{x}}^2-\frac{1}{{\mathrm{x}}^2})}=2^{2(1-\frac{1}{{\mathrm{x}}^2})}$$$$\Rightarrow \mathrm{x}=1\mathrm{and}1-\frac{1}{{\mathrm{x}}^2}=0$$

Commented by chux last updated on 09/Jul/17

$$\mathrm{wow}!!\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}.$$

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