Question

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

Answer 1

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

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

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

We can write it as $x=\frac{1}{3\sqrt{2}}\sqrt{4-x}$

$\Rightarrow 3\sqrt{2}x=\sqrt{4-x}$

Squaring on both sides, we get $18x^2=4-x$

$18x^2+x-4=0$

Solving for $x$ we get $x=-\frac{18}{36},\frac{16}{36}$

But as $x$ can't be negative $\therefore x=\frac{16}{36}=\frac{4}{9}$

Put value of $x$ in $\left(1\right)$,

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

$6-2=4$

Hence, the value is $4$.