Question

The value of
$\displaystyle 6 + \log_{\frac{3}{2}} \left( \frac{1}{3\sqrt{2}} \sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\cdots}}} \right)$​

Answer 1

$HEYA \: \\ \\ 6 + log_{(3 \div 2)}((1 \div 3 \sqrt{2} )( \sqrt{4 - 1 \div 3 \sqrt{2} } ) \times \ \sqrt{4 - 1 \div 3 \sqrt{2} }) .. \infty \\ \\ put \: \: \: \: (4 - 1 \div 3 \sqrt{2} ) = z \\ \\ 6 + log_{(3 \div 2)}((1 \div 3 \sqrt{2}) \times \sqrt{z \sqrt{z \sqrt{z \sqrt{z} } } } ... \infty ) \\ \\ 6 + log_{(3 \div 2)}(1 \div 3 \sqrt{2} ) + log_{(3 \div 2)}( \sqrt{z \sqrt{z \sqrt{z} } }... \infty ) \\ \\ 6 + log_{(3 \div 2)}(1 \div 3 \sqrt{2} ) + (1 \div 2) log_{(3 \div 2}(z) \\ + (1 \div 4) log_{(3 \div 2)}(z) + (1 \div 8) log_{(3 \div 2) }(z) ... \infty \\ \\ \\ 6 + log_{3 \div 2}(1 \div 3 \sqrt{2} ) + log_{(3 \div 2)}(z) ((1 \div 2) + (1 \div 4) + (1 \div 8) + ... \infty ) \\ \\ \\ 6 + log_{(3 \div 2)}(1 \div 3 \sqrt{2} ) + log_{(3 \div 2)}(1) \\ becoz \: (1 \div 2) + (1 \div 4) + (1 \div 8) + .. \infty \\ are \: in \: g.p \\ \\ 6 + log_{(3 \div 2)}(1 \div 3 \sqrt{2} ) \\ \\ 6 - log_{(3 \div 2)}(3 \sqrt{2} )$

Answer 2

Answer:

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