The value of 6 + log 3 2 ( 1 3 2 4 - 1 3 2 4 - 1 3 2 4 - 1 3 2 ? )

Question

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

Smart Achievers · Accepted Answer

(1)

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

$t^{2} = 4 - \frac{1}{3 \sqrt{2}} t \Rightarrow t^{2} + \frac{1}{3 \sqrt{2}} t - 4 = 0 \Rightarrow t = \frac{8}{3 \sqrt{2}}$

$The required answer$

$= 6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \times \frac{8}{3 \sqrt{2}}) = 6 + \log_{\frac{3}{2}} \frac{4}{9} = 6 - 2 = 4$

Solution

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

1 4

2 3

3 2

4 1

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Solution

Q. The value of 6+log32(1324-1324-1324-132⋯)

1 4

2 3

3 2

4 1

Ans. (1) Let t=4-1324-132⋯ t2=4-132t⇒t2+132t-4=0⇒t=832 The required answer =6+log32(132×832)=6+log3249=6-2=4

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