Question

If ${\log }_{2\sqrt{2}}\sqrt{m}+{\log }_2\sqrt[3]{3\frac{1}{6}}=\frac{2}{3}$, then the value of $m$ is

Answer 1

${\log }_{2\sqrt{2}}\sqrt{m}+{\log }_2\sqrt[3]{3\frac{1}{6}}=\frac{2}{3}\Rightarrow {\log }_{2^{3/2}}{m}^{1/2}+\frac{1}{3}{\log }_2\frac{1}{6}=\frac{2}{3}\Rightarrow {\log }_{8}m+{\log }_8\frac{1}{6}=\frac{2}{3}\Rightarrow {\log }_8\left(\frac{m}{6}\right)=\frac{2}{3}\Rightarrow \frac{m}{6}=8^{2/3}=4\therefore m=24$