Question

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

• A. $4$
• B. $5$
• C. $6$
• D. $9$

Answer 1

Answer: (A)

$4$

Let

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

Hence

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

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

$y=\frac{-1\pm \sqrt{1+4\left(72\right)}}{6\sqrt{2}}$

$y=\frac{-1\pm 17}{6\sqrt{2}}$

Since input of logarithmic functions cannot be negative.

Therefore

$y=\frac{16}{6\sqrt{2}}$

$=\frac{4\sqrt{2}}{3}$

Hence

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

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

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

$=\frac{4}{9}$

Therefore

$6+lo{g}_{\frac{3}{2}}\left(\frac{4}{9}\right)$

$=6+lo{g}_{\frac{3}{2}}(\frac{3}{2}{)}^{-2}$

$=6-2$

$=4$