Question Number 122821 by bemath last updated on 19/Nov/20
$$\:{Given}\:\begin{cases}{{x}^{\mathrm{ln}\:{x}+\mathrm{ln}\:{y}} \:=\:\mathrm{5}}\\{{y}^{\mathrm{ln}\:{x}+\mathrm{ln}\:{y}} \:=\:\mathrm{2}}\end{cases} \\ $$$${then}\:\begin{cases}{{x}=?}\\{{y}=?}\end{cases} \\ $$
Answered by liberty last updated on 20/Nov/20
![{ ((ln (xy).ln (x) = ln (5) )),((ln (xy).ln (y) = ln (2))) :} → ((ln (5))/(ln (x))) = ((ln (2))/(ln (y))) ⇒ ln (y)=((ln (2))/(ln (5))).ln (x) ⇒ { ln (x)+((ln (2))/(ln (5))).ln (x)}.{((ln (2))/(ln (5))).ln (x)}=ln (2) ln^2 (x) [ 1+((ln (2))/(ln (5))) ].((ln (2))/(ln (5))) = ln (2) ln^2 (x) [ ((ln (10).ln (2))/(ln^2 (5))) ]= ln (2) ln^2 (x) = ((ln^2 (5))/(ln (10))) ⇒ln (x) = ± ((ln (5))/( (√(ln (10))))) ⇒x = e^(± (((ln (5))/( (√(ln (10))))))) .](https://www.tinkutara.com/question/Q122831.png)
$$\:\begin{cases}{\mathrm{ln}\:\left({xy}\right).\mathrm{ln}\:\left({x}\right)\:=\:\mathrm{ln}\:\left(\mathrm{5}\right)\:}\\{\mathrm{ln}\:\left({xy}\right).\mathrm{ln}\:\left({y}\right)\:=\:\mathrm{ln}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\rightarrow\:\frac{\mathrm{ln}\:\left(\mathrm{5}\right)}{\mathrm{ln}\:\left({x}\right)}\:=\:\frac{\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:\left({y}\right)}\:\Rightarrow\:\mathrm{ln}\:\left({y}\right)=\frac{\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:\left(\mathrm{5}\right)}.\mathrm{ln}\:\left({x}\right) \\ $$$$\Rightarrow\:\left\{\:\mathrm{ln}\:\left({x}\right)+\frac{\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:\left(\mathrm{5}\right)}.\mathrm{ln}\:\left({x}\right)\right\}.\left\{\frac{\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:\left(\mathrm{5}\right)}.\mathrm{ln}\:\left({x}\right)\right\}=\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)\:\left[\:\mathrm{1}+\frac{\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:\left(\mathrm{5}\right)}\:\right].\frac{\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:\left(\mathrm{5}\right)}\:=\:\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)\:\left[\:\frac{\mathrm{ln}\:\left(\mathrm{10}\right).\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{5}\right)}\:\right]=\:\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{5}\right)}{\mathrm{ln}\:\left(\mathrm{10}\right)}\:\Rightarrow\mathrm{ln}\:\left({x}\right)\:=\:\pm\:\frac{\mathrm{ln}\:\left(\mathrm{5}\right)}{\:\sqrt{\mathrm{ln}\:\left(\mathrm{10}\right)}} \\ $$$$\Rightarrow{x}\:=\:{e}^{\pm\:\left(\frac{\mathrm{ln}\:\left(\mathrm{5}\right)}{\:\sqrt{\mathrm{ln}\:\left(\mathrm{10}\right)}}\right)} . \\ $$