please help me to solve this sistem: { ((log_2 (x+2y)−log


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Question Number 53740 by F_Nongue last updated on 25/Jan/19
$please help me to solve this sistem: { ((log_2 (x+2y)−log_3 (x−2y)=2)),((x^2 −4y^2 =4)) :}$
$p l e a s e h e l p m e t o s o l v e t h i s$ $s i s t e m :$ ${\begin{cases} {l o g}_{2} (x + 2 y) - {l o g}_{3} (x - 2 y) = 2 \\ x^{2} - 4 y^{2} = 4 \end{cases}$
	Answered by Smail last updated on 25/Jan/19

	$x^{2} - 4 y^{2} = 4 \Rightarrow x = \underline{+} \sqrt{4 + 4 y^{2}} = \underline{+} 2 \sqrt{1 + y^{2}}$ ${l o g}_{2} (x + 2 y) - {l o g}_{3} (x - 2 y) = 2$ $\frac{l n (x + 2 y)}{l n 2} - \frac{l n (x - 2 y)}{l n 3} = 2$ $\frac{l n (2 \sqrt{1 + y^{2}} + 2 y)}{l n 2} - \frac{l n (2 \sqrt{1 + y^{2}} - 2 y)}{l n 3} = 2$ $1 + \frac{l n (\sqrt{1 + y^{2}} + y)}{l n 2} - \frac{l n 2}{l n 3} - \frac{l n (\sqrt{1 + y^{2}} - y)}{l n 3} = 2$ $1 + \frac{l n (\sqrt{1 + y^{2}} + y)}{l n 2} - \frac{l n 2}{l n 3} - \frac{l n (\frac{1 + y^{2} - y^{2}}{\sqrt{1 + y^{2}} + y})}{l n 3} = 2$ $1 + \frac{{s i n h}^{- 1} (y)}{l n 2} - {l o g}_{3} 2 + \frac{{s i n h}^{- 1} (y)}{l n 3} = 2$ ${s i n h}^{- 1} (y) (\frac{l n 3 + l n 2}{l n 2 \times l n 3}) = 1 + \frac{l n 2}{l n 3} = \frac{l n 3 + l n 2}{l n 3}$ ${s i n h}^{- 1} (y) = l n 2$ $y = s i n h (l n 2) = \frac{e^{l n 2} - e^{- l n 2}}{2}$ $y = \frac{3}{4}$ $x = 2 \sqrt{1 + y^{2}} = 2 \sqrt{1 + \frac{9}{16}}$ $x = \frac{5}{2}$
	Commented by F_Nongue last updated on 27/Jan/19

	$p l e a s e s i r c a n y o u e x p l a i n y o u r s t e p s ?$
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