The value of 6 + log 3 2 ( 1 3 2 ? 4 - 1 3 2 ? 4 - 1 3 2 ? 4 - 1 3 2 ? ) is [2012]

Question

The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}})$ is [2012]

Smart Achievers · Accepted Answer

(4)

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

$\Rightarrow x = \frac{1}{3 \sqrt{2}} \sqrt{4 - x} \Rightarrow 3 \sqrt{2} x = \sqrt{4 - x}$

$\Rightarrow 18 x^{2} = 4 - x \Rightarrow 18 x^{2} + x - 4 = 0$

$\Rightarrow (9 x - 4) (2 x + 1) = 0 \Rightarrow x = \frac{4}{9} or x = - \frac{1}{2}$

(Not possible because log is not defined for negative value)

$∴$ $6 + \log_{\frac{3}{2}} (\frac{4}{9}) = 6 + \log_{\frac{3}{2}} {(\frac{3}{2})}^{- 2} = 6 - 2 = 4$

Solution

Q.
The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}})$ is [2012]

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Solution

Q. The value of 6+log32(132 4-132 4-132 4-132⋯) is [2012]

Ans. (4) Let x=1324-1324-1324-132⋯ ⇒x=1324-x⇒32x=4-x ⇒18x2=4-x⇒18x2+x-4=0 ⇒(9x-4)(2x+1)=0⇒x=49 or x=-12 (Not possible because log is not defined for negative value) ∴ 6+log32(49)=6+log32(32)-2=6-2=4

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Q.
The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}})$ is [2012]