The value of 6+log _{frac{3}{2}}left(frac{1}{ | vedclass

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Question

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{

Vedclass · Accepted Answer

$4$

Vedclass · Answer

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 $

$t=\frac{-1 \pm \sqrt{1+4 	imes 3 \sqrt{2} 	imes 12 \sqrt{2}}}{2 	imes 3 \sqrt{2}}=\frac{-1 \pm 17}{2 	imes 3 \sqrt{2}}$

$t=\frac{16}{6 \sqrt{2}}, \frac{-18}{6 \sqrt{2}}$

$t =\frac{8}{3 \sqrt{2}}, \frac{-3}{\sqrt{2}}$ and $\frac{-3}{\sqrt{2}}$ is rejected

$	ext { so } 6+\log _{3 / 2}\left(\frac{1}{3 \sqrt{2}} 	imes \frac{8}{3 \sqrt{2}}ight)=6+\log _{3 / 2}\left(\frac{4}{9}ight)=6+\log _{3 / 2}\left(\left(\frac{2}{3}ight)^2ight)=6-2=4$

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

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