Question

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

Answer 1

Given $6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}....}}}\right)$
Let $y=\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}....}}$

$\Rightarrow y=\sqrt{4-\frac{y}{3\sqrt{2}}}$

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

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

$\therefore y=\frac{-1\pm \sqrt{1-4\times 3\sqrt{12}\times -12\sqrt{2}}}{2\times 3\sqrt{2}}$

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

$=\frac{-1\pm \sqrt{289}}{6\sqrt{2}}$

$=\frac{-1\pm 17}{6\sqrt{2}}$

$=\frac{16}{6\sqrt{2}},\frac{-18}{6\sqrt{2}}$

Negative value of $y=\frac{-18}{6\sqrt{2}}$ is not valid since $\log$ is not defined for negative value.

$\therefore y=\frac{16}{6\sqrt{2}}=\frac{8}{3\sqrt{2}}$

Now, Consider, $6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{2\sqrt{2}}\sqrt{4-\frac{1}{2\sqrt{2}}\sqrt{4-\frac{1}{2\sqrt{2}}....}}}\right)$

$=6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}y\right)$

$=6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\frac{8}{3\sqrt{2}}\right)$ where $y=\frac{8}{3\sqrt{2}}$

$=6+{\log }_{\frac{3}{2}}\left(\frac{8}{9\times 2}\right)$

$=6+{\log }_{\frac{3}{2}}\left(\frac{4}{9}\right)$

$=6+{\log }_{\frac{3}{2}}\left(\frac{2^2}{3^2}\right)$

$=6+{\log }_{\frac{3}{2}}{\left(\frac{3}{2}\right)}^{-2}$

$=6-2=4$