Question

The value of $3+{\log }_{9/4}\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\cdots \cdots \infty }}}$ is

Answer 1

Let $\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\cdots \cdots \infty }}}=y,y>0$
$\Rightarrow 4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\cdots \cdots \infty }}=y^2$
$\Rightarrow 4-\frac{1}{3\sqrt{2}}\cdot y=y^2$
$\Rightarrow y^2+\frac{y}{3\sqrt{2}}-4=0$
$\Rightarrow 3\sqrt{2}{y}^2+y-12\sqrt{2}=0$
$\Rightarrow y=\frac{-1\pm \sqrt{1+288}}{6\sqrt{2}}$
$\Rightarrow y=\frac{-1+17}{6\sqrt{2}}\text{or}y=\frac{-1-17}{6\sqrt{2}}\left(\text{rejected}\right)$
$\Rightarrow y=\frac{8}{3\sqrt{2}}$

$\therefore 3+{\log }_{{\left(3/2\right)}^2}(\frac{1}{3\sqrt{2}}\cdot \frac{8}{3\sqrt{2}})$
$=3+\frac{1}{2}{\log }_{3/2}\left(\frac{4}{9}\right)$
$=3+\frac{1}{2}{\log }_{3/2}{\left(\frac{2}{3}\right)}^2$
$=3+\frac{1}{2}\cdot 2{\log }_{3/2}\left(\frac{2}{3}\right)$
$=3-\frac{1}{2}\cdot 2{\log }_{3/2}\left(\frac{3}{2}\right)$
$=3-1=2$