Question

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

Answer 1

Let $\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}..............\infty }}}=x$

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

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

$12\sqrt{2}-x={3\sqrt{2}x}^2$

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

${3\sqrt{2}x}^2+9x-8x-12\sqrt{2}=0$

${3x(\sqrt{2}x}^{}+3)-4\sqrt{2}(\sqrt{2}x+3)=0$

$(3x-4\sqrt{2}\left)\right(\sqrt{2}x+3)=0$

$x=\frac{4\sqrt{2}}{3},\frac{-3}{\sqrt{2}}$

x can not be negative

so, $x=\frac{4\sqrt{2}}{3}$

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

$=6+{\log }_{\frac{3}{2}}(\frac{1}{3\sqrt{2}}\times \frac{4\sqrt{2}}{3})$

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

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

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

$=6-2$

$=4$