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Question Number 84459 by jagoll last updated on 13/Mar/20
{ ((log_(10) (x)+((log_(10) (x)+8log_(10) (y))/(log_(10) ^2 (x)+log_(10) ^2 (y)))=3)),((log_(10) (y)+((8log_(10) (x)−log_(10) (y))/(log_(10) ^2 (x)+log_(10) ^2 (y)))=0)) :} find x & y
$$\begin{cases}{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\frac{\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{8log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{3}}\\{\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)+\frac{\mathrm{8log}_{\mathrm{10}} \left(\mathrm{x}\right)−\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)}{\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{log}_{\mathrm{10}} ^{\mathrm{2}} \left(\mathrm{y}\right)}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{x}\:\&\:\mathrm{y} \\ $$
Commented by john santu last updated on 13/Mar/20
let log_(10) (x) = a & log_(10) (y)=b (1) a+((a+8b)/(a^2 +b^2 )) = 2 ∧ (2) b +((8a−b)/(a^2 +b^2 )) = 0 b(a+((a+8b)/(a^2 +b^2 ))) = 2b (1) a(b+((8a−b)/(a^2 +b^2 ))) = 0 (2) ⇒2ab +((8a^2 +8b^2 )/(a^2 +b^2 )) = 2b
$$\mathrm{let}\:\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}\right)\:=\:\mathrm{a}\:\&\:\mathrm{log}_{\mathrm{10}} \left(\mathrm{y}\right)=\mathrm{b} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{a}+\frac{\mathrm{a}+\mathrm{8b}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{2}\:\wedge\:\left(\mathrm{2}\right)\:\mathrm{b}\:+\frac{\mathrm{8a}−\mathrm{b}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\mathrm{b}\left(\mathrm{a}+\frac{\mathrm{a}+\mathrm{8b}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\right)\:=\:\mathrm{2b}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{a}\left(\mathrm{b}+\frac{\mathrm{8a}−\mathrm{b}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\right)\:=\:\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2ab}\:+\frac{\mathrm{8a}^{\mathrm{2}} +\mathrm{8b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{2b} \\ $$
Commented by jagoll last updated on 13/Mar/20
2ab+8=2b ⇒ 4 = b−ab b = (4/(1−a)) ∧ a^2 b+b^3 +8a−b=0 thank you sir. i have idea to solving this problem
$$\mathrm{2ab}+\mathrm{8}=\mathrm{2b}\:\Rightarrow\:\mathrm{4}\:=\:\mathrm{b}−\mathrm{ab} \\ $$$$\mathrm{b}\:=\:\frac{\mathrm{4}}{\mathrm{1}−\mathrm{a}}\:\wedge\:\mathrm{a}^{\mathrm{2}} \mathrm{b}+\mathrm{b}^{\mathrm{3}} +\mathrm{8a}−\mathrm{b}=\mathrm{0} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{have}\:\mathrm{idea}\:\mathrm{to}\: \\ $$$$\mathrm{solving}\:\mathrm{this}\:\mathrm{problem} \\ $$
Answered by MJS last updated on 13/Mar/20
log_(10) x =u∧log_(10) y =v v=pu∧p, u≠0 { ((u−3+((8p+1)/((p^2 +1)u))=0)),((pu−((p−8)/((p^2 +1)u))=0)) :} { ((u^2 −3u+((8p+1)/(p^2 +1))=0)),((u^2 −((p−8)/(p(p^2 +1)))=0)) :} (2)−(1) ⇒ u=((2(4p^2 +p−4))/(3p(p^2 +1))) insert in (1) or (2) ⇒ p^4 +((104)/(55))p^3 −((133)/(55))p^2 +(8/(11))p+((64)/(55))=0 (p^2 +((4−(13+(√(1049))))/(55))p+((123+(√(1049)))/(110)))(p^2 +((4(13−(√(1049))))/(55))p+((123−(√(1049)))/(110)))=0 we can exactly solve this
$$\mathrm{log}_{\mathrm{10}} \:{x}\:={u}\wedge\mathrm{log}_{\mathrm{10}} \:{y}\:={v} \\ $$$${v}={pu}\wedge{p},\:{u}\neq\mathrm{0} \\ $$$$\begin{cases}{{u}−\mathrm{3}+\frac{\mathrm{8}{p}+\mathrm{1}}{\left({p}^{\mathrm{2}} +\mathrm{1}\right){u}}=\mathrm{0}}\\{{pu}−\frac{{p}−\mathrm{8}}{\left({p}^{\mathrm{2}} +\mathrm{1}\right){u}}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{u}^{\mathrm{2}} −\mathrm{3}{u}+\frac{\mathrm{8}{p}+\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0}}\\{{u}^{\mathrm{2}} −\frac{{p}−\mathrm{8}}{{p}\left({p}^{\mathrm{2}} +\mathrm{1}\right)}=\mathrm{0}}\end{cases} \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\:\Rightarrow\:{u}=\frac{\mathrm{2}\left(\mathrm{4}{p}^{\mathrm{2}} +{p}−\mathrm{4}\right)}{\mathrm{3}{p}\left({p}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${p}^{\mathrm{4}} +\frac{\mathrm{104}}{\mathrm{55}}{p}^{\mathrm{3}} −\frac{\mathrm{133}}{\mathrm{55}}{p}^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{11}}{p}+\frac{\mathrm{64}}{\mathrm{55}}=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +\frac{\mathrm{4}−\left(\mathrm{13}+\sqrt{\mathrm{1049}}\right)}{\mathrm{55}}{p}+\frac{\mathrm{123}+\sqrt{\mathrm{1049}}}{\mathrm{110}}\right)\left({p}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{13}−\sqrt{\mathrm{1049}}\right)}{\mathrm{55}}{p}+\frac{\mathrm{123}−\sqrt{\mathrm{1049}}}{\mathrm{110}}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{this} \\ $$

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