Discussion Forum

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

Answer:

Explanation:

Concept involved 

 use of infinite series

 i.e  if y=$\sqrt{x\sqrt{x\sqrt{x....\infty}}}\Rightarrow y=\sqrt{xy}$

 sol.

$6+\log_{3/2}$

$\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}}}}\right)$

 Let  $\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{....}}}= y$

$\therefore$   $y=\sqrt{4-\frac{1}{3\sqrt{2}}}y$

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

 $\Rightarrow$  $3 \sqrt{2} y^{2}+y-12\sqrt{2}=0$

 $\therefore$   $y=\frac{-1\pm 17}{6\sqrt{2}}$  or  $y=\frac{8}{3\sqrt{2}}$

 Now, 

$6+\log_{3/2}\left(\frac{1}{3\sqrt{2}}.y\right)=6+\log_{3/2}\left(\frac{1}{3\sqrt{2}}.\frac{8}{3\sqrt{2}}\right)$

 =$6+\log_{3/2} \left(\frac{4}{9}\right)=6+\log _{3/2}\left(\frac{3}{2}\right)^{-2}$

=$6-2\log_{3/2} \left(\frac{3}{2}\right)=4$