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

Answer 1

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

$\Rightarrow L=\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}}\sqrt{4-\frac{1}{3\sqrt{2}}}}....$

$L=\sqrt{4-\frac{1}{3\sqrt{2}}}L$

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

$\Rightarrow 3\sqrt{2}{L}^2+L-12\sqrt{2}=0$

$L=\frac{-1\pm \sqrt{1+288}}{6\sqrt{2}}$

$L=\frac{-1+\sqrt{289}}{6\sqrt{2}}\Rightarrow L=\frac{-1+17}{6\sqrt{2}}$

$L=\frac{-1+17}{6\sqrt{2}}=\frac{16}{6\sqrt{2}}=\frac{4\sqrt{2}}{3}$

$\Rightarrow 6+lo{g}_{3/2}(\frac{1}{3\sqrt{2}}\times \frac{4\sqrt{2}}{3})$

$\Rightarrow 6+lo{g}_{3/2}\left(\frac{4}{9}\right)=6+lo{g}_{3/2}{\left(\frac{3}{2}\right)}^{-2}$

$\Rightarrow 6+(-2)lo{g}_{3/2}\left(3/2\right)$

$\Rightarrow 6-2=4$