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The value of
$6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}} \cdots}}\right) \text { is }$
$6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}} \cdots}}\right) \text { is }$
Solution:
1749 Upvotes
Verified Answer
The correct answer is:
4
Let $x=\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}}} \ldots \ldots$
$\Rightarrow \quad x=\frac{1}{3 \sqrt{2}} \sqrt{4-x}$
$\Rightarrow \quad 3 \sqrt{2} x=\sqrt{4-x}$
$\begin{array}{l}
\Rightarrow \quad 18 x^{2}=4-x \\
\Rightarrow \quad 18 x^{2}+x-4=0 \\
\Rightarrow \quad(9 x-4)(2 x+1)=0 \\
\Rightarrow \quad x=\frac{4}{9} \text { or } x=-\frac{1}{2}
\end{array}$
(Not possible because log is not define for -ve value)
$\begin{aligned}
\therefore \quad & 6+\log _{\frac{3}{2}}\left(\frac{4}{9}\right)=6+\log _{\frac{3}{2}}\left(\frac{3}{2}\right)^{-2} \\
&=6-2=4
\end{aligned}$
$\Rightarrow \quad x=\frac{1}{3 \sqrt{2}} \sqrt{4-x}$
$\Rightarrow \quad 3 \sqrt{2} x=\sqrt{4-x}$
$\begin{array}{l}
\Rightarrow \quad 18 x^{2}=4-x \\
\Rightarrow \quad 18 x^{2}+x-4=0 \\
\Rightarrow \quad(9 x-4)(2 x+1)=0 \\
\Rightarrow \quad x=\frac{4}{9} \text { or } x=-\frac{1}{2}
\end{array}$
(Not possible because log is not define for -ve value)
$\begin{aligned}
\therefore \quad & 6+\log _{\frac{3}{2}}\left(\frac{4}{9}\right)=6+\log _{\frac{3}{2}}\left(\frac{3}{2}\right)^{-2} \\
&=6-2=4
\end{aligned}$
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