$$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$$

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

$$\therefore$$ $$y = \sqrt {4 - {1 \over {3\sqrt 2 }}y} $$

$$ \Rightarrow {y^2} + {1 \over {3\sqrt 2 }}y - 4 = 0$$

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

$$\therefore$$ $$y = {{ - 1 \pm 17} \over {6\sqrt 2 }}$$ or $$y = {8 \over {3\sqrt 2 }}$$

Now,

$$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.y} \right) = 6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.{8 \over {3\sqrt 2 }}} \right)$$

$$ = 6 + {\log _{{3 \over 2}}}\left( {{4 \over 9}} \right) = 6 + {\log _{{3 \over 2}}}{\left( {{3 \over 2}} \right)^{ - 2}}$$

$$ = 6 - 2.{\log _{{3 \over 2}}}\left( {{3 \over 2}} \right) = 4$$