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 that,

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

Let us consider $y=\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{(4-\frac{1}{3\sqrt{2)}}}\sqrt{4-\frac{1}{3\sqrt{2}}}}$

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

squaring on both sides

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

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

solving for $y$

$y=\frac{-1+\sqrt{1-(4\times 3\sqrt{2}\times -12\sqrt{2})}}{2\times 3\sqrt{2}}$

$y=\frac{-1+\sqrt{289}}{6\sqrt{2}}=\frac{17-1}{6\sqrt{2}}=\frac{16}{6\sqrt{2}}$

$y=\frac{8}{3\sqrt{2}}$

Substituting in $\left(1\right)$

$6+lo{g}_{\frac{3}{2}}[\frac{1}{3\sqrt{2}}\times \frac{8}{3\sqrt{2}}]=6+lo{g}_{\frac{3}{2}}^{\frac{4}{9}}$

$=6-lo{g}_{\frac{3}{2}}^{\frac{9}{4}}$

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

$=6-2lo{g}_{\frac{3}{2}}^{\frac{3}{2}}$

$=6-2$

$6+lo{g}_{\frac{3}{2}}[\frac{1}{3\sqrt{2}}\times \frac{8}{3\sqrt{2}}]$$=4$

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