Question

The value of
6 + $ log_{ 3/2} \Bigg ( \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} } ...}} \Bigg ) $

Answer 1

PLAN Use of infinite series
i, e., if y = $\sqrt{ x \sqrt {x} \sqrt x .... \infty } \Rightarrow y = \sqrt{ xy}$.
Given, 6 + $ log_\frac{ 3}{2} \Bigg ( \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} } ...}} \Bigg ) $
Let $ \sqrt{ 4 - \frac{ 1}{ 3 \sqrt 2} \sqrt{4 - \frac{ 1}{ 3 \sqrt 2} \sqrt ..... = y $
$\therefore y = \sqrt{ 4 - \frac{1}{ 3 \sqrt 2} y } $
$\Rightarrow y^2 + \frac{1}{ 3 \sqrt 2} y - 4 = 0 $
$\Rightarrow 3 \sqrt 2 y^2 + y - 12 \sqrt 2 = 0 $
$\therefore $ $ $ $ y = \frac{ - 1 \pm 17 }{ 6 \sqrt 2} $
or $ y = \frac{ 8}{ 43 \sqrt 2}$
$ = 6 + log_{ \frac{ 3}{ 2}} \bigg( \frac[ 4}{ 9} \bigg) = 6 + log_{ \frac{ 3}{ 2}} \bigg( \frac[ 3}{ 2} \bigg)^{ - 2}$
= 6 - 2 - $ log_{ \frac{ 3}{ 2}} \bigg( \frac[ 3}{ 2} \bigg) = 4 $