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

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(A) Let $t = \sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\dots}}}$.Then $t = \sqrt{4-\frac{1}{3\sqrt{2}}t}$.Squaring both sides,we g

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$4$

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(A) Let $t = \sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\dots}}}$.Then $t = \sqrt{4-\frac{1}{3\sqrt{2}}t}$.Squaring both sides,we get $4-\frac{1}{3\sqrt{2}}t = t^2$,which implies $t^2 + \frac{1}{3\sqrt{2}}t - 4 = 0$.Multiplying by $3\sqrt{2}$,we get $3\sqrt{2}t^2 + t - 12\sqrt{2} = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have $t = \frac{-1 \pm \sqrt{1 - 4(3\sqrt{2})(-12\sqrt{2})}}{2(3\sqrt{2})} = \frac{-1 \pm \sqrt{1 + 288}}{6\sqrt{2}} = \frac{-1 \pm 17}{6\sqrt{2}}$.Since $t > 0$,we take $t = \frac{16}{6\sqrt{2}} = \frac{8}{3\sqrt{2}}$.The expression becomes $6 + \log_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}} \cdot \frac{8}{3\sqrt{2}}\right) = 6 + \log_{\frac{3}{2}}\left(\frac{8}{18}\right) = 6 + \log_{\frac{3}{2}}\left(\frac{4}{9}\right)$.Since $\frac{4}{9} = (\frac{2}{3})^2 = (\frac{3}{2})^{-2}$,the expression is $6 + \log_{\frac{3}{2}}\left((\frac{3}{2})^{-2}\right) = 6 - 2 = 4$.

The value of $6+\log_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\dots}}}\right)$ is

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