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

The value of 6 + log_(3/2) ( 1/3√2 √4 - 1/3√2 √4 - 1/3√2 √4 - 1/3√2 ·s ) is $\text{Given: } \dfrac{1}{3\sqrt{2}}\sqrt{4 - \dfrac{1}{3\sqrt{2}}\sqrt{4 - \dfrac{1}{3\sqrt{2}}\sqrt{4 - \cdots}}}$ $\text{To find: } 6 + \log_{\frac{3}{2}}\left(\ \cdot\ \right)$ $\text{Plan:}$ $1.\ \text{call the infinite radical } y$ $2.\ \text{self-similarity} \Rightarrow \text{quadratic} \Rightarrow \text{solve} \Rightarrow \log$ $y = \dfrac{1}{3\sqrt{2}}\sqrt{4 - \dfrac{1}{3\sqrt{2}}\sqrt{4 - \cdots}}$ $\text{The inner part repeats } y:$ $y = \dfrac{1}{3\sqrt{2}}\sqrt{4 - y}$ $y^2 = \dfrac{1}{18}\,(4 - y)$ $18y^2 = 4 - y$ $18y^2 + y - 4 = 0$ $\Delta = 1^2 - 4(18)(-4) = 1 + 288$ $\Delta = 289 = 17^2$ $y = \dfrac{-1 \pm \sqrt{289}}{36} = \dfrac{-1 \pm 17}{36}$ $y = \dfrac{16}{36} = \dfrac{4}{9}\quad \text{or}\quad y = -\dfrac{1}{2}$ $y > 0 \Rightarrow y = \dfrac{4}{9}\ \ (-\tfrac{1}{2}\ \text{rejected})$ $\dfrac{4}{9} = \left(\dfrac{2}{3}\right)^{2} = \left(\dfrac{3}{2}\right)^{-2}$ $\log_{3/2}\!\left(\dfrac{4}{9}\right) = \log_{3/2}\!\left(\left(\dfrac{3}{2}\right)^{-2}\right) = -2$ $6 + \log_{3/2}\!\left(\dfrac{4}{9}\right) = 6 + (-2) = 4$ $\boxed{\text{Value} = 4}$