The value of 6 + \left(log\right)_{\frac{3}{2}} \left(\right. \frac{1}{3 \sqrt{2}} \textrm{ } \sqrt{4 \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{1}{3 \sqrt{2}} \textrm{ } \sqrt{4 \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{1}{3 \sqrt{2}} \textrm{ } \sqrt{4 \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{1}{3 \sqrt{2}}} . .....}} \left.\right) is

Engineering

Mathematics

Properties of Logarithm

Question

The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}}} . .....}})$ is

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$Let x = \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} . .....}}}$

$x = \frac{1}{3 \sqrt{2}} \sqrt{4 - x} \Rightarrow 18 x^{2} = 4 - x$

$\Rightarrow$ 18x² + x – 4 = 0

$x = \frac{- 1 \pm \sqrt{1 + 18 \times 4 \times 4}}{36} = \frac{- 1 \pm 17}{36} = \frac{16}{36} = \frac{4}{9} = {(\frac{2}{3})}^{2}$

$Hence, 6 + \log_{\frac{3}{2}} {(\frac{2}{3})}^{2} = 6 - 2 = 4$