A simple radical expression!

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

It is easy to see, that A > 0 .

Then:

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

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

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

Solving the quadratic equation above for A > 0 , we get:

$A = \frac {- \frac {1}{ 3 \sqrt {2} } + \sqrt { \frac {289}{18} }}{2} = \frac {-1 + 17}{ 6 \sqrt {2} } = \frac {8}{ 3 \sqrt {2} }$

Now, our original expression can be written as:

$6 + \log _{ \frac {3}{2} }{( \frac {1}{ 3 \sqrt {2} } A) } = 6 + \log _{ \frac {3}{2} }{( \frac {1}{ 3 \sqrt {2} } × \frac {8}{ 3 \sqrt {2} } ) } = 6 + \log _{ \frac {3}{2} }{( \frac {4}{9}) } = 6 + (-2) = \boxed {4}$

Consider $\sqrt { 4-\dfrac { 1 }{ 3\sqrt { 2 } } \sqrt { 4-\dfrac { 1 }{ 3\sqrt { 2 } } \sqrt { 4-\dfrac { 1 }{ 3\sqrt { 2 } } \cdots } } } = x$

$\Longrightarrow x^{2} = 4-\dfrac { 1 }{ 3\sqrt { 2 }} \sqrt { 4-\dfrac { 1 }{ 3\sqrt { 2 } } \sqrt { 4-\dfrac { 1 }{ 3\sqrt { 2 } } \cdots } }$

$\Longrightarrow x^{2} = 4-\dfrac{1}{3\sqrt{2}}x$

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

$\Longrightarrow x=\dfrac{-1\pm\sqrt{1-4\left(-12\sqrt{2}\right)\left(3\sqrt{2}\right)}}{6\sqrt{2}}$

$\Longrightarrow x=\dfrac{-1\pm17}{6\sqrt{2}}$

$\Longrightarrow x=\dfrac{8}{3\sqrt{2}}$ or $x=\dfrac{-3}{\sqrt{2}}$

$\Longrightarrow x=\dfrac{4\sqrt{2}}{3}$ or $x=\dfrac{-3\sqrt{2}}{2}$

Out of the following two solutions we need $x$ to be positive to carry logarithmic operation on it.

$\therefore x=\dfrac{4\sqrt{2}}{3}$

Now, $\dfrac{1}{3\sqrt{2}}\times\dfrac{4\sqrt{2}}{3}=\dfrac{4}{9}$

Hence, $\log_{{3}/{2}}{\dfrac{4}{9}}=-2$

And so, $6+\left(-2\right)=\large\boxed{4}$