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Question-197388

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Question Number 197388 by cortano12 last updated on 16/Sep/23
Answered by MM42 last updated on 16/Sep/23
x=1 & y=1
$${x}=\mathrm{1}\:\:\&\:\:{y}=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Sep/23
Find x and y { ((9x^2 −4y^2 =5............(i))),((log_5 (3x+2y)−log_5 (3x−2y)=1...(ii) )) :} (i)⇒(3x+2y)(3x−2y)=5 ⇒log_5 (3x+2y)+log_5 (3x−2y)=1...(iii) (ii)+(iii): 2log_5 (3x+2y)=2 log_5 (3x+2y)=1 3x+2y=5........A (iii)−(ii): 2log_5 (3x−2y)=0 3x−2y=5^0 =1.....B A+B: 6x=6⇒x=1 A−B: 4y=4⇒y=1
$${Find}\:{x}\:{and}\:{y} \\ $$$$\begin{cases}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{5}…………\left({i}\right)}\\{\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}+\mathrm{2}{y}\right)−\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}−\mathrm{2}{y}\right)=\mathrm{1}…\left({ii}\right)\:\:}\end{cases} \\ $$$$\left({i}\right)\Rightarrow\left(\mathrm{3}{x}+\mathrm{2}{y}\right)\left(\mathrm{3}{x}−\mathrm{2}{y}\right)=\mathrm{5} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}+\mathrm{2}{y}\right)+\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}−\mathrm{2}{y}\right)=\mathrm{1}…\left({iii}\right) \\ $$$$\left({ii}\right)+\left({iii}\right): \\ $$$$\:\:\mathrm{2log}_{\mathrm{5}} \left(\mathrm{3}{x}+\mathrm{2}{y}\right)=\mathrm{2} \\ $$$$\:\:\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}+\mathrm{2}{y}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{3}{x}+\mathrm{2}{y}=\mathrm{5}……..{A} \\ $$$$\left({iii}\right)−\left({ii}\right): \\ $$$$\mathrm{2log}_{\mathrm{5}} \left(\mathrm{3}{x}−\mathrm{2}{y}\right)=\mathrm{0} \\ $$$$\:\:\:\mathrm{3}{x}−\mathrm{2}{y}=\mathrm{5}^{\mathrm{0}} =\mathrm{1}…..{B} \\ $$$${A}+{B}: \\ $$$$\mathrm{6}{x}=\mathrm{6}\Rightarrow{x}=\mathrm{1} \\ $$$${A}−{B}: \\ $$$$\mathrm{4}{y}=\mathrm{4}\Rightarrow{y}=\mathrm{1} \\ $$
Commented by cortano12 last updated on 16/Sep/23
nice
$$\mathrm{nice}\: \\ $$
Answered by Rasheed.Sindhi last updated on 16/Sep/23
x,y=? { ((9x^2 −4y^2 =5.......(i))),((log_5 (3x+2y)−log_5 (3x−2y)=1...(ii) )) :} (ii)⇒log_5 (((3x+2y)/(3x−2y)))=1 ((3x+2y)/(3x−2y))=5 ((3x+2y)/(3x−2y))+1=6 ((6x)/(3x−2y))=6 x=3x−2y 2x−2y=0⇒x=y (i)⇒9x^2 −4x^2 =5 x^2 =1=y^2 x=y=1 ∨ x=y=−1( invalid)
$${x},{y}=? \\ $$$$\:\begin{cases}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{5}…….\left({i}\right)}\\{\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}+\mathrm{2}{y}\right)−\mathrm{log}_{\mathrm{5}} \left(\mathrm{3}{x}−\mathrm{2}{y}\right)=\mathrm{1}…\left({ii}\right)\:\:}\end{cases} \\ $$$$\left({ii}\right)\Rightarrow\mathrm{log}_{\mathrm{5}} \left(\frac{\mathrm{3}{x}+\mathrm{2}{y}}{\mathrm{3}{x}−\mathrm{2}{y}}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{3}{x}+\mathrm{2}{y}}{\mathrm{3}{x}−\mathrm{2}{y}}=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{3}{x}+\mathrm{2}{y}}{\mathrm{3}{x}−\mathrm{2}{y}}+\mathrm{1}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{6}{x}}{\mathrm{3}{x}−\mathrm{2}{y}}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{3}{x}−\mathrm{2}{y} \\ $$$$\:\:\:\:\:\:\mathrm{2}{x}−\mathrm{2}{y}=\mathrm{0}\Rightarrow{x}={y} \\ $$$$\left({i}\right)\Rightarrow\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} =\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} =\mathrm{1}={y}^{\mathrm{2}} \\ $$$${x}={y}=\mathrm{1}\:\vee\:{x}={y}=−\mathrm{1}\left(\:{invalid}\right) \\ $$$$\:\:\:\:\:\: \\ $$
Answered by MM42 last updated on 16/Sep/23
(3x−2y)(3x+2y)=5⇒3x−2y=(5/(3x+2y)) 1−log_5 (3x−2y)−log_3 (3x−2y)=1 −((log(3x−2y))/(log5)) −((log(3x−2y))/(log3))=0 log(3x−2)(−(1/(log5))−(1/(log3)))=0 ⇒3x−2y=1⇒3x+2y=5 ⇒x=y=1 ✓
$$\left(\mathrm{3}{x}−\mathrm{2}{y}\right)\left(\mathrm{3}{x}+\mathrm{2}{y}\right)=\mathrm{5}\Rightarrow\mathrm{3}{x}−\mathrm{2}{y}=\frac{\mathrm{5}}{\mathrm{3}{x}+\mathrm{2}{y}} \\ $$$$\mathrm{1}−{log}_{\mathrm{5}} \left(\mathrm{3}{x}−\mathrm{2}{y}\right)−{log}_{\mathrm{3}} \left(\mathrm{3}{x}−\mathrm{2}{y}\right)=\mathrm{1} \\ $$$$−\frac{{log}\left(\mathrm{3}{x}−\mathrm{2}{y}\right)}{{log}\mathrm{5}}\:−\frac{{log}\left(\mathrm{3}{x}−\mathrm{2}{y}\right)}{{log}\mathrm{3}}=\mathrm{0} \\ $$$${log}\left(\mathrm{3}{x}−\mathrm{2}\right)\left(−\frac{\mathrm{1}}{{log}\mathrm{5}}−\frac{\mathrm{1}}{{log}\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}−\mathrm{2}{y}=\mathrm{1}\Rightarrow\mathrm{3}{x}+\mathrm{2}{y}=\mathrm{5} \\ $$$$\Rightarrow{x}={y}=\mathrm{1}\:\checkmark \\ $$$$ \\ $$

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