Question

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

• A. $6$
• B. $8$
• C. $2$
• D. $4$

Answer 1

The correct option is D $4$

Let
$t=\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\dots \dots }}}$
Then $t=\frac{1}{3\sqrt{2}}\sqrt{4-t}$
$\Rightarrow t^2=\frac{1}{18}(4-t)$
$\Rightarrow 18t^2+t-4=0$
$\Rightarrow t=\frac{4}{9}\text{or}t=-\frac{1}{2}\left(\text{Not possible}\right)$
$\therefore t=\frac{4}{9}$

Now, $y=6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\dots \dots }}}\right)$
$\Rightarrow y=6+{\log }_{\frac{3}{2}}\left(\frac{4}{9}\right)$
$\Rightarrow y=6+{\log }_{\frac{3}{2}}{\left(\frac{3}{2}\right)}^{-2}\Rightarrow y=6-2=4$