Answer:
4
Step-by-step explanation:
Given,
[tex]6+log\frac{3}{2} (\frac{1}{3\sqrt{2} } \sqrt{4-\frac{1}{3\sqrt{2} }\sqrt{4-\frac{1}{3\sqrt{2} } ...} }[/tex]
Let,
[tex]x= \sqrt{4-\frac{1}{3\sqrt{2} }\sqrt{4-\frac{1}{3\sqrt{2} }\\[/tex]
By this we get
[tex]x=\sqrt{4-\frac{1}{3\sqrt{2} }(x) } }[/tex]
On squaring both sides,
[tex]x^{2} =4-\frac{1}{3\sqrt{2} }(x) } }\\\\x^{2} -4+\frac{x}{3\sqrt{2} } =0\\\\3\sqrt{2} x^{2} -12\sqrt{2} +x=0\\\\x=\frac{-1+\sqrt{1-(-12\sqrt{2})*(3\sqrt{2})*4 } }{2*3\sqrt{2} } \\\\x=\frac{-1+\sqrt{289} }{6\sqrt{2} } \\\\x=\frac{-1+17}{6\sqrt{2} } \\\\x=\frac{8}{3\sqrt{2} }[/tex]
Now,
[tex]6+log\frac{3}{2} [\frac{1}{3\sqrt{2} } *\frac{8}{3\sqrt{2} } ]+log\frac{3}{2} *[\frac{8}{9*2} ]\\\\6+log\frac{3}{2}(\frac{4}{9} )\\\\6-log\frac{3}{2}(\frac{9}{4} )\\\\6-log\frac{3}{2} (\frac{3}{2})^2 \\\\6-2=4[/tex]
[tex]\sqrt{4 - \dfrac1{3\sqrt2} \sqrt{4 - \dfrac1{3\sqrt2} \sqrt{4 - \dfrac1{3\sqrt2} \sqrt{\cdots}}}}[/tex]
Starting from the identity
[tex](x - y)^2 = x^2 - 2xy + y^2[/tex]
take the positive square root on both sides.
[tex]x - y = \sqrt{x^2 - 2xy + y^2}[/tex]
Note that we must have [tex]x\ge y[/tex]. Rewrite the radicand and substitute [tex]x-y[/tex].
[tex]x - y = \sqrt{x^2 - xy - y (x - y)} \\\\ ~~~~ = \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y (x - y)}} \\\\ ~~~~ = \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y (x - y)}}} \\\\ ~~~~ \vdots \\\\ ~~~~ = \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y \sqrt{x^2 - xy - y \sqrt{\cdots}}}}[/tex]
Let [tex]y=\frac1{3\sqrt2}[/tex]. Solve for [tex]x[/tex].
[tex]x^2 - \dfrac x{3\sqrt2} = 4 \\\\ x^2 - \dfrac x{3\sqrt2} + \dfrac1{72} = \dfrac{289}{72} \\\\ \left(x - \dfrac1{6\sqrt2}\right)^2 = \dfrac{289}{72} \\\\ x - \dfrac1{6\sqrt2} = \pm \dfrac{17}{6\sqrt2} \\\\ x = \dfrac{18}{6\sqrt2} \text{ or } x = -\dfrac{16}{6\sqrt2} \\\\ x = \dfrac3{\sqrt2} \text{ or } x = -\dfrac8{3\sqrt2}[/tex]
Take the positive solution to ensure [tex]x>y[/tex]. Then the infinitely nested root expression in the logarithm converges to
[tex]x - y = \dfrac3{\sqrt2} - \dfrac1{3\sqrt2} = \dfrac{4\sqrt2}3[/tex]
and the overall expression has a value of
[tex]6 + \log_{\frac32} \left(\dfrac1{3\sqrt2} \times \dfrac{4\sqrt2}3\right) = 6 + \log_{\frac32} \left(\dfrac49\right) \\\\ ~~~~ = 6 + \log_{\frac32} \left(\dfrac23\right)^2 \\\\ ~~~~ = 6 - 2 \log_{\frac32} \left(\dfrac32\right) \\\\ ~~~~ = 6 - 2 = \boxed{4}[/tex]
