Question

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

Answer 1

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

squaring on both sides

$x^2=\frac{4-x}{18}$

$18x^2+x-4=0⟹x=\frac{4}{9}$

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