Question Number 160976 by alf123 last updated on 10/Dec/21
Answered by TheSupreme last updated on 10/Dec/21
![(x/y)=s 2^(sy) +3^y =21 2^(s/2) =3 → s = 2 ((ln(3))/(ln(2))) 3^(2y) +3^y =21 t^2 +3t=21 t=((−3±(√(9+84)))/2)=((−3+(√(93)))/2) y=ln(3)[(((√(93))−3)/2)] x=((ln^2 (3))/(ln(2)))[(√(93))−3] (1/x)+(1/y)=(1/( (√(93))−3))[((ln(2))/(ln^2 (3)))+(1/(ln(3)))]](Q160977.png)
$$\frac{{x}}{{y}}={s} \\ $$$$\mathrm{2}^{{sy}} +\mathrm{3}^{{y}} =\mathrm{21} \\ $$$$\mathrm{2}^{\frac{{s}}{\mathrm{2}}} =\mathrm{3}\:\rightarrow\:{s}\:=\:\mathrm{2}\:\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)} \\ $$$$\mathrm{3}^{\mathrm{2}{y}} +\mathrm{3}^{{y}} =\mathrm{21} \\ $$$${t}^{\mathrm{2}} +\mathrm{3}{t}=\mathrm{21} \\ $$$${t}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{84}}}{\mathrm{2}}=\frac{−\mathrm{3}+\sqrt{\mathrm{93}}}{\mathrm{2}} \\ $$$${y}={ln}\left(\mathrm{3}\right)\left[\frac{\sqrt{\mathrm{93}}−\mathrm{3}}{\mathrm{2}}\right] \\ $$$${x}=\frac{{ln}^{\mathrm{2}} \left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)}\left[\sqrt{\mathrm{93}}−\mathrm{3}\right] \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{93}}−\mathrm{3}}\left[\frac{{ln}\left(\mathrm{2}\right)}{{ln}^{\mathrm{2}} \left(\mathrm{3}\right)}+\frac{\mathrm{1}}{{ln}\left(\mathrm{3}\right)}\right] \\ $$$$ \\ $$
Answered by FelipeLz last updated on 10/Dec/21
![{ ((2^x −3^y = 21)),(((√2^(x/y) ) = 3)) :} (√2^(x/y) ) = 3 ⇒ (2^(x/y) )^(1/2) = 3 ⇒ 2^(x/y) = 3^2 ⇒ 2^x = 3^(2y) 2^x −3^y = 21 ⇒ 3^(2y) −3^y −21 = 0 3^y = u ⇒ u^2 −u−21 = 0 u = ((1±(√(1+4∙21)))/2) = ((1±(√(85)))/2) 3^y = ((1±(√(85)))/2) ⇒ y = log_3 (((1±(√(85)))/2)) = ((ln(1±(√(85)))−ln(2))/(ln(3))) 2^x = 3^(2y) ⇒ xln(2) = 2yln(3) ⇒ xln(2) = 2ln(3)[((ln(1±(√(85)))−ln(2))/(ln(3)))] ⇒ x = 2[((ln(1±(√(85))))/(ln(2)))−1] = 2log_2 (((1±(√(85)))/2)) { ((x = 2log_2 (((1±(√(85)))/2)))),((y = log_3 (((1±(√(85)))/2)))) :} ((1±(√(85)))/2) = a (1/x)+(1/y) = (1/(2log_2 (a))) + (1/(log_3 (a))) = (1/2)log_a (2)+log_a (3) = log_a ((√2))+log_a (3) = log_a (3(√2)) (1/x)+(1/y) = log_((((1±(√(85)))/2))) (3(√2))](Q160978.png)
$$\begin{cases}{\mathrm{2}^{{x}} −\mathrm{3}^{{y}} \:=\:\mathrm{21}}\\{\sqrt{\mathrm{2}^{\frac{{x}}{{y}}} }\:=\:\mathrm{3}}\end{cases} \\ $$$$\sqrt{\mathrm{2}^{\frac{{x}}{{y}}} }\:=\:\mathrm{3}\:\Rightarrow\:\left(\mathrm{2}^{\frac{{x}}{{y}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{3}\:\Rightarrow\:\mathrm{2}^{\frac{{x}}{{y}}} \:=\:\mathrm{3}^{\mathrm{2}} \:\Rightarrow\:\mathrm{2}^{{x}} \:=\:\mathrm{3}^{\mathrm{2}{y}} \: \\ $$$$\mathrm{2}^{{x}} −\mathrm{3}^{{y}} \:=\:\mathrm{21}\:\Rightarrow\:\mathrm{3}^{\mathrm{2}{y}} −\mathrm{3}^{{y}} −\mathrm{21}\:=\:\mathrm{0} \\ $$$$\mathrm{3}^{{y}} \:=\:{u}\:\Rightarrow\:{u}^{\mathrm{2}} −{u}−\mathrm{21}\:=\:\mathrm{0} \\ $$$${u}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\centerdot\mathrm{21}}}{\mathrm{2}}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}} \\ $$$$\mathrm{3}^{{y}} \:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\:\Rightarrow\:{y}\:=\:\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{ln}\left(\mathrm{1}\pm\sqrt{\mathrm{85}}\right)−\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{3}\right)} \\ $$$$\mathrm{2}^{{x}} \:=\:\mathrm{3}^{\mathrm{2}{y}} \:\Rightarrow\:{x}\mathrm{ln}\left(\mathrm{2}\right)\:=\:\mathrm{2}{y}\mathrm{ln}\left(\mathrm{3}\right)\:\Rightarrow\:{x}\mathrm{ln}\left(\mathrm{2}\right)\:=\:\mathrm{2ln}\left(\mathrm{3}\right)\left[\frac{\mathrm{ln}\left(\mathrm{1}\pm\sqrt{\mathrm{85}}\right)−\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{3}\right)}\right]\:\Rightarrow\:{x}\:=\:\mathrm{2}\left[\frac{\mathrm{ln}\left(\mathrm{1}\pm\sqrt{\mathrm{85}}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}−\mathrm{1}\right]\:=\:\mathrm{2log}_{\mathrm{2}} \left(\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\right) \\ $$$$\begin{cases}{{x}\:=\:\mathrm{2log}_{\mathrm{2}} \left(\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\right)}\\{{y}\:=\:\mathrm{log}_{\mathrm{3}} \left(\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\right)}\end{cases} \\ $$$$\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\:=\:{a} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:=\:\frac{\mathrm{1}}{\mathrm{2log}_{\mathrm{2}} \left({a}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{3}} \left({a}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{{a}} \left(\mathrm{2}\right)+\mathrm{log}_{{a}} \left(\mathrm{3}\right)\:=\:\mathrm{log}_{{a}} \left(\sqrt{\mathrm{2}}\right)+\mathrm{log}_{{a}} \left(\mathrm{3}\right)\:=\:\mathrm{log}_{{a}} \left(\mathrm{3}\sqrt{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:=\:\mathrm{log}_{\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{85}}}{\mathrm{2}}\right)} \left(\mathrm{3}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$