Question

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

Answer 1

$y=6+lo{g}_{3/2}\left(\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}}}\right)$

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

$=x$

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

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

$\Rightarrow 3\sqrt{2}x=\sqrt{4-x}$

$\Rightarrow (3\sqrt{2}x{)}^2=4-x$

$\Rightarrow 18x^2+x-4=0$

$\Rightarrow 18x^2+9x-8x-4=0$

$\Rightarrow 9x(2x+1)-4(2x+1)=0$

$\Rightarrow (9x-4)(2x+1)=0$

$\Rightarrow x=-\frac{1}{2}$ or $x=\frac{4}{9}$

but $x=\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}}$ can not be negative

$\Rightarrow x=\frac{4}{9}$

$\Rightarrow y=6+lo{g}_{3/2}\left(x\right)=6+lo{g}_{3/2}\left(\frac{4}{9}\right)$

$=6+lo{g}_{3/2}(3/2{)}^{-2}$

$=6+(-2)\left(lo{g}_{3/2}3/2\right)$

$=6-2\left(1\right)=6-2$

$=4$