The problem is to evaluate the expression $6 + \log_{\frac{3}{2}} \left\{ \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \cdots }}} \right\}$.

1. Problem Description

The problem is to evaluate the expression $6 + \log_{\frac{3}{2}} \left\{ \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \cdots }}} \right\}$.

2. Solution Steps

Let $x = \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \cdots }}}$

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

Squaring both sides, we get $x^2 = 4 - \frac{1}{3\sqrt{2}} x$.

Rearranging the terms, we have $x^2 + \frac{1}{3\sqrt{2}} x - 4 = 0$.

Multiplying by $3\sqrt{2}$, we get $3\sqrt{2} x^2 + x - 12\sqrt{2} = 0$.

Using the quadratic formula, we have

$x = \frac{-1 \pm \sqrt{1^2 - 4(3\sqrt{2})(-12\sqrt{2})}}{2(3\sqrt{2})} = \frac{-1 \pm \sqrt{1 + 4(3)(12)(2)}}{6\sqrt{2}} = \frac{-1 \pm \sqrt{1 + 288}}{6\sqrt{2}} = \frac{-1 \pm \sqrt{289}}{6\sqrt{2}} = \frac{-1 \pm 17}{6\sqrt{2}}$.

Since $x$ must be positive, we take the positive root, so $x = \frac{-1 + 17}{6\sqrt{2}} = \frac{16}{6\sqrt{2}} = \frac{8}{3\sqrt{2}}$.

Then the expression inside the logarithm is $\frac{1}{3\sqrt{2}} \cdot \frac{8}{3\sqrt{2}} = \frac{8}{9(2)} = \frac{8}{18} = \frac{4}{9}$.

Now we have $6 + \log_{\frac{3}{2}} \left( \frac{4}{9} \right) = 6 + \log_{\frac{3}{2}} \left( \left( \frac{2}{3} \right)^2 \right) = 6 + 2 \log_{\frac{3}{2}} \left( \frac{2}{3} \right) = 6 + 2 \log_{\frac{3}{2}} \left( \left( \frac{3}{2} \right)^{-1} \right) = 6 + 2(-1) \log_{\frac{3}{2}} \left( \frac{3}{2} \right) = 6 - 2(1) = 6 - 2 = 4$.

3. Final Answer

4