Given that log_4 (y−1)+log_4 ((x/y))=m and log_2 (y+1)−log


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Question Number 170468 by MathsFan last updated on 24/May/22

$G i v e n t h a t {l o g}_{4} (y - 1) + {l o g}_{4} (\frac{x}{y}) = m$ $a n d {l o g}_{2} (y + 1) - {l o g}_{2} x = m - 1,$ $s h o w t h a t y^{2} = 1 - 8^{m}$
	Answered by cortano1 last updated on 24/May/22
	${ ((log _4 (y−1)+log _4 ((x/y))=m)),((log _2 (y+1)−log _2 x=m−1)) :} { ((log _2 (((y−1)/y))+log _2 x = 2m)),((log _2 (y+1)−log _2 x=m−1)) :} (1)+(2) ⇒log _2 (((y^2 −1)/y))=3m−1 ⇒((y^2 −1)/y) = (8^m /2)$
	${\begin{cases} \log_{4} (y - 1) + \log_{4} (\frac{x}{y}) = m \\ \log_{2} (y + 1) - \log_{2} x = m - 1 \end{cases}$ ${\begin{cases} \log_{2} (\frac{y - 1}{y}) + \log_{2} x = 2 m \\ \log_{2} (y + 1) - \log_{2} x = m - 1 \end{cases}$ $(1) + (2)$ $\Rightarrow \log_{2} (\frac{y^{2} - 1}{y}) = 3 m - 1$ $\Rightarrow \frac{y^{2} - 1}{y} = \frac{8^{m}}{2}$
	Commented by MathsFan last updated on 24/May/22

	$t h a n k s$
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