Answer:
Option A
Explanation:
Concept involved
use of infinite series
i.e if y=$\sqrt{x\sqrt{x\sqrt{x....\infty}}}\Rightarrow y=\sqrt{xy}$
sol.
$6+\log_{3/2}$
$\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}}}}\right)$
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 $y=\frac{8}{3\sqrt{2}}$
Now,
$6+\log_{3/2}\left(\frac{1}{3\sqrt{2}}.y\right)=6+\log_{3/2}\left(\frac{1}{3\sqrt{2}}.\frac{8}{3\sqrt{2}}\right)$
=$6+\log_{3/2} \left(\frac{4}{9}\right)=6+\log _{3/2}\left(\frac{3}{2}\right)^{-2}$
=$6-2\log_{3/2} \left(\frac{3}{2}\right)=4$
