Question

Solve :
$log_{2x}(2) + log_{4}(2x) = - \frac{3}{2}$
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Answer 1

Answer:

$x = {2}^{ - 2} \: or \: {2}^{ - 3}$

✌️Hope it will help you.✌️

Answer 2

Answer:

$= > \frac{1}{1 + log {}^{x} _ {2}} + \frac{ log_{2x}}{ \frac{2}{2} } = \frac{ - 3}{2} \\ \\ or \: \: \: = > \frac{1}{1 + log {}^{x} _{2} } + \frac{1 + log \: x}{ \frac{2}{2} } = \frac{ - 3}{2}$

Let 1 + log2 x = y

$= > \frac{1}{y} + \frac{y}{2} = \frac{ - 3}{2} \\ = > 2 + y {}^{2} + 3y = 0 \\ = > y = - 1 \: \: \: or \: \: \: - 2 \\ = > 1 + log_{2} \: x = - 1 \: \: \: or \: \: - 2 \\ = > log_{2} \: x = - 2 \: \: \: or \: \: \: - 3 \\ = > x = 2 {}^{ - 2} \: \: \: or \: \: \: 2 {}^{ - 3}$