Question

Solve the logarithm question In JEE

$6 + log_{ \frac{3}{2} }( \frac{1}{3 \sqrt{2} } \sqrt{4 - \frac{1}{3 \sqrt{2} } \sqrt{4 - \frac{1}{3 \sqrt{2} } \sqrt{4 - \frac{1}{3 \sqrt{2} }..... } } } ) \: is$
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Answer 1

Step-by-step explanation:

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Answer 2

$\blue{\bold{\underline{\underline{\boxed{answer : }}}}} \:$

$let \: \: y = \sqrt{4 - \frac{1}{3 \sqrt{2} }y } \\ \\ {y}^{2} = 4 - \frac{y}{3 \sqrt{2} } \\ \\ 3 \sqrt{2} {y}^{2} + y - 12 \sqrt{2 } = 0 \\ \\ y = - 1 + - \sqrt{1 + 4 \times 12 \sqrt{2} \times 3 \sqrt{2} } \: \div 6 \sqrt{2} \\ \\ y = \frac{ - 1 + - 17}{6 \sqrt{2} } = \frac{8}{3 \sqrt{2} } \\ \\ 6 + log_{ \frac{3}{2} }( \frac{1}{3 \sqrt{2} } \times \frac{8}{3 \sqrt{2} } ) \\ \\ 6 + log_{ \frac{3}{2} }( \frac{4}{9} ) \\ \\ log_{ \frac{3}{2} }( \frac{4}{9} ) = y \\ \\ ( \frac{3}{2}) ^{y} = \frac{4}{9} = ( \frac{2}{3} ) ^{2} = ( \frac{3}{2} ) ^{ - 2} \\ \\ y = - 2 \\ \\ 6 + ( - 2) = 4 \\ \\ therefore \: value \: of \: the \: question \: is \: \boxed{\underline{4}} \\ \\ mark \: it \: as \: brainliest \: and \: \\ follow \: me \\ \\ have \: a \: good \: day$