Evaluation of Logarithmic Expression with Nested Square Roots

To find the exact value of the expression $E = 6 + \log_{\frac{3}{2}} \left( \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \cdots}}} \right)$

we first define the nested radical term as $x$: $x = \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \cdots}}}$

Since the expression inside the square root repeats infinitely, we can rewrite $x$ self-referentially as: $x = \frac{1}{3\sqrt{2}} \sqrt{4 - x}$

Since $x$ represents a principal square root multiplied by a positive constant, we must have $x > 0$.

Squaring both sides of the equation yields: $x^2 = \left( \frac{1}{3\sqrt{2}} \right)^2 (4 - x)$ $x^2 = \frac{1}{18} (4 - x)$

Rearranging this into a standard quadratic equation: $18x^2 + x - 4 = 0$

We factor the quadratic equation: $18x^2 + 9x - 8x - 4 = 0$ $9x(2x + 1) - 4(2x + 1) = 0$ $(9x - 4)(2x + 1) = 0$

This gives two possible roots for $x$: $x = \frac{4}{9} \quad \text{or} \quad x = -\frac{1}{2}$

Since $x > 0$, we reject the negative root, giving: $x = \frac{4}{9}$

Now, substituting $x$ back into the original expression $E$: $E = 6 + \log_{\frac{3}{2}} (x) = 6 + \log_{\frac{3}{2}} \left( \frac{4}{9} \right)$

Notice that: $\frac{4}{9} = \left( \frac{9}{4} \right)^{-1} = \left( \left( \frac{3}{2} \right)^2 \right)^{-1} = \left( \frac{3}{2} \right)^{-2}$

Using the logarithm property $\log_b (b^k) = k$: $\log_{\frac{3}{2}} \left( \frac{4}{9} \right) = \log_{\frac{3}{2}} \left( \frac{3}{2} \right)^{-2} = -2$

Finally, we calculate the total value: $E = 6 + (-2) = 4$