Question

The value of given expression is $6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}.....}}\right)$

• A. $4$
• B. $5$
• C. $7$
• D. None of these

Answer 1

The correct option is A

$4$

Explanation for the correct option:

Find the required value.

Given:$6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}.....}}\right)$

Let $y=6+{\log }_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}.....}}\right)$

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

$$\begin{gathered} \Rightarrow x=\frac{1}{3\sqrt{2}}\sqrt{4-x} \\ \Rightarrow 3\sqrt{2}x=\sqrt{4-x} \\ \Rightarrow {\left(3\sqrt{2}x\right)}^2={\left(\sqrt{4-x}\right)}^2 \\ \Rightarrow 18x^2=4-x \\ \Rightarrow 18x^2+x-4=0 \\ \Rightarrow 18x^2+9x-8x-4=0 \\ \Rightarrow 9x\left(2x+1\right)-4(2x+1)=0 \\ \Rightarrow \left(9x-4\right)\left(2x+1\right)=0 \\ \Rightarrow x=\frac{4}{9}orx=\frac{-1}{2} \end{gathered}$$

But $x=\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}.....}}$ can not be negative

$\therefore x=\frac{4}{9}$

Now,

$y=6+{\log }_{\frac{3}{2}}\left(x\right)$

$$\begin{gathered} =6+{\log }_{\frac{3}{2}}\left(\frac{4}{9}\right) \\ =6+{\log }_{\frac{3}{2}}{\left(\frac{3}{2}\right)}^{-2} \\ =6+(-2)\left({\log }_{\frac{3}{2}}\left(\frac{3}{2}\right)\right) \\ =6-2\left(1\right) \\ =6-2 \\ =4 \end{gathered}$$

Hence, option A is correct .