Math, asked by SharmaShivam, 1 year ago

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)

Answers

Answered by Anonymous
3

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} )

Answered by Anonymous
2

Answer:

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