Question

The value of ${\log }_{(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}})}{2}^9$ is

Answer 1

Let $L={\log }_{(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}})}{2}^9$
and $S=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}$
$=\sqrt{(\sqrt{2}+1{)}^2}+\sqrt{(\sqrt{2}-1{)}^2}$
$=|\sqrt{2}+1|+|\sqrt{2}-1|$
$=2\sqrt{2}$

$L={\log }_{2\sqrt{2}}{2}^9$
$=\frac{{\log }_2{2}^9}{{\log }_{2}2\sqrt{2}}=\frac{9}{3/2}=\frac{2\times 9}{3}=6$