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$log^2(2y\;+\;3)\;-log_24 = log_2(y-2)$
for expressing 2 as log<sub>2</sub>4
$log_2(\frac{2y+3}{4}) \; =\;log_2(y\;-\;2)$
$\frac{2y\;+\;3}{4}\;=\frac{y\;-\;2}{1}$
$2y\;+\;3\;=4y\;-\;8; 11y\; =\; 2y$
y = 5.5
johnmulu answered the question on March 8, 2017 at 04:48
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