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Question Number 197388 by cortano12 last updated on 16/Sep/23

	Answered by MM42 last updated on 16/Sep/23

	$x = 1 & y = 1$
	Answered by Rasheed.Sindhi last updated on 16/Sep/23
	$Find x and y { ((9x^2 −4y^2 =5............(i))),((log_5 (3x+2y)−log_5 (3x−2y)=1...(ii) )) :} (i)⇒(3x+2y)(3x−2y)=5 ⇒log_5 (3x+2y)+log_5 (3x−2y)=1...(iii) (ii)+(iii): 2log_5 (3x+2y)=2 log_5 (3x+2y)=1 3x+2y=5........A (iii)−(ii): 2log_5 (3x−2y)=0 3x−2y=5^0 =1.....B A+B: 6x=6⇒x=1 A−B: 4y=4⇒y=1$
	$F i n d x a n d y$ ${\begin{cases} 9 x^{2} - 4 y^{2} = 5 . . . . . . . . . . . . (i) \\ \log_{5} (3 x + 2 y) - \log_{5} (3 x - 2 y) = 1 . . . (i i) \end{cases}$ $(i) \Rightarrow (3 x + 2 y) (3 x - 2 y) = 5$ $\Rightarrow \log_{5} (3 x + 2 y) + \log_{5} (3 x - 2 y) = 1 . . . (i i i)$ $(i i) + (i i i) :$ ${2 \log}_{5} (3 x + 2 y) = 2$ $\log_{5} (3 x + 2 y) = 1$ $3 x + 2 y = 5 . . . . . . . . A$ $(i i i) - (i i) :$ ${2 \log}_{5} (3 x - 2 y) = 0$ $3 x - 2 y = 5^{0} = 1 . . . . . B$ $A + B :$ $6 x = 6 \Rightarrow x = 1$ $A - B :$ $4 y = 4 \Rightarrow y = 1$
	Commented by cortano12 last updated on 16/Sep/23

	$nice$
	Answered by Rasheed.Sindhi last updated on 16/Sep/23
	$x,y=? { ((9x^2 −4y^2 =5.......(i))),((log_5 (3x+2y)−log_5 (3x−2y)=1...(ii) )) :} (ii)⇒log_5 (((3x+2y)/(3x−2y)))=1 ((3x+2y)/(3x−2y))=5 ((3x+2y)/(3x−2y))+1=6 ((6x)/(3x−2y))=6 x=3x−2y 2x−2y=0⇒x=y (i)⇒9x^2 −4x^2 =5 x^2 =1=y^2 x=y=1 ∨ x=y=−1( invalid)$
	$x, y = ?$ ${\begin{cases} 9 x^{2} - 4 y^{2} = 5 . . . . . . . (i) \\ \log_{5} (3 x + 2 y) - \log_{5} (3 x - 2 y) = 1 . . . (i i) \end{cases}$ $(i i) \Rightarrow \log_{5} (\frac{3 x + 2 y}{3 x - 2 y}) = 1$ $\frac{3 x + 2 y}{3 x - 2 y} = 5$ $\frac{3 x + 2 y}{3 x - 2 y} + 1 = 6$ $\frac{6 x}{3 x - 2 y} = 6$ $x = 3 x - 2 y$ $2 x - 2 y = 0 \Rightarrow x = y$ $(i) \Rightarrow 9 x^{2} - 4 x^{2} = 5$ $x^{2} = 1 = y^{2}$ $x = y = 1 \lor x = y = - 1 (i n v a l i d)$
	Answered by MM42 last updated on 16/Sep/23

	$(3 x - 2 y) (3 x + 2 y) = 5 \Rightarrow 3 x - 2 y = \frac{5}{3 x + 2 y}$ $1 - {l o g}_{5} (3 x - 2 y) - {l o g}_{3} (3 x - 2 y) = 1$ $- \frac{l o g (3 x - 2 y)}{l o g 5} - \frac{l o g (3 x - 2 y)}{l o g 3} = 0$ $l o g (3 x - 2) (- \frac{1}{l o g 5} - \frac{1}{l o g 3}) = 0$ $\Rightarrow 3 x - 2 y = 1 \Rightarrow 3 x + 2 y = 5$ $\Rightarrow x = y = 1 ✓$
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