Question

plz solve this equation.
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Answer 1

S O L U T I O N :

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$\sf : \implies \: log_{x} 2.log_{2x} 2 = log_{4x} 2 \\ \\ \\ \sf : \implies \: \frac{1}{log_{2} x} . \frac{1}{log_{2} 2x } = \frac{1}{log_{2} 4x} \\ \\ \\ \sf : \implies \: \frac{1}{log_{2} x}.\frac{1}{(log_{2} 2 + log_{2} x)} = \frac{1}{(log_{2} 4 +log_{2} x) }$

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$\sf \: Let \: the \: value \: of \: log_{ 2} x \: be \: \bold{t}.$

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$\sf : \implies \: \left( \dfrac{1}{t} \right)\left( \dfrac{1}{1 + t} \right) = \left( \dfrac{1}{ 2 + t} \right) \\ \\ \\ \sf : \implies \:t \: (1 + t) = (2 + t) \\ \\ \\ \sf : \implies \: {t}^{2} + t = 2 + t \\ \\ \\ \sf : \implies \: {t}^{2} = 2 \\ \\ \\ \sf : \implies \:t = + \sqrt{2} , - \sqrt{2}$

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$\sf \: Now, \: let's \: put \: the \: value \: of \: \bold{t}$

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$\sf : \implies \: log_{ 2} x = \sqrt{2}, - \sqrt{2} \\ \\ \\ \sf : \implies \:x = {2}^{ \sqrt{2} } \: , {2}^{ - \sqrt{2} }$

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$\sf \therefore\underline{Hence, \: the \: answer \: is \: \bold{{2}^{ \sqrt{2} } } \: and \: \bold{{2}^{ - \sqrt{2} }} }$