The value of 6+log3/21324-1324-1324-132⋯ is

Let S=1324-1324-1324-132⋯∞⇒ S=1324-S or (32S)2=4-S ⇒ 18S2+S-4=0 ⇒ (9S-4)(2S+1)=0 ∴ 9S-4=0 or S=49=32-2 [∵2S+1≠0] ⇒ log3/2S=-2⇒6+log3/2S=6-2=4

The value of 6+log3/21324-1324-1324-132⋯ is

Let S=1324-1324-1324-132⋯∞⇒ S=1324-S or (32S)2=4-S ⇒ 18S2+S-4=0 ⇒ (9S-4)(2S+1)=0 ∴ 9S-4=0 or S=49=32-2 [∵2S+1≠0] ⇒ log3/2S=-2⇒6+log3/2S=6-2=4

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

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$5$

$2$

$1$

$4$

answer is D.

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Detailed Solution

Let $S = \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots \infty}}}$

$\Rightarrow S = \frac{1}{3 \sqrt{2}} \sqrt{4 - S} or (3 \sqrt{2} S)^{2} = 4 - S \Rightarrow 18 S^{2} + S - 4 = 0 \Rightarrow (9 S - 4) (2 S + 1) = 0 ∴ 9 S - 4 = 0 or S = \frac{4}{9} = {(\frac{3}{2})}^{- 2} [∵ 2 S + 1 \neq 0] \Rightarrow \log_{3 / 2} S = - 2 \Rightarrow 6 + \log_{3 / 2} S = 6 - 2 = 4$