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The value of $6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \ldots}}}\right)$ is
$4$
$5$
$3$
$1$
Solution
Let $\sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}} \ldots \ldots . .= t$
$\sqrt{4-\frac{1}{3 \sqrt{2}} t}=t $
$4-\frac{1}{3 \sqrt{2}} t=t^2 \Rightarrow $
$t^2+\frac{1}{3 \sqrt{2}} t-4=0 \Rightarrow 3 \sqrt{2} t^2+t-12 \sqrt{2}=0 $
$t=\frac{-1 \pm \sqrt{1+4 \times 3 \sqrt{2} \times 12 \sqrt{2}}}{2 \times 3 \sqrt{2}}=\frac{-1 \pm 17}{2 \times 3 \sqrt{2}}$
$t=\frac{16}{6 \sqrt{2}}, \frac{-18}{6 \sqrt{2}}$
$t =\frac{8}{3 \sqrt{2}}, \frac{-3}{\sqrt{2}}$ and $\frac{-3}{\sqrt{2}}$ is rejected
$\text { so } 6+\log _{3 / 2}\left(\frac{1}{3 \sqrt{2}} \times \frac{8}{3 \sqrt{2}}\right)=6+\log _{3 / 2}\left(\frac{4}{9}\right)=6+\log _{3 / 2}\left(\left(\frac{2}{3}\right)^2\right)=6-2=4$
