$\text{ If }{\log }_8\mathrm{m}+{\log }_8\left(\frac{1}{6}\right)=\frac{2}{3}\text{, then }\mathrm{m}\text{ is equal to: }$

$$\begin{gathered} {\log }_8\mathrm{m}+{\log }_8\left(\frac{1}{6}\right)=\frac{2}{3} \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that}, {\log }_{\mathrm{n}}(\mathrm{x}\times \mathrm{y})={\log }_{\mathrm{n}}\mathrm{x}+{\log }_{\mathrm{n}}\mathrm{y}\text{ where }\mathrm{n}\ne 1 \\ \Rightarrow {\log }_8\mathrm{m}+{\log }_8\left(\frac{1}{6}\right)={\log }_8\left(\frac{\mathrm{m}}{6}\right)=\frac{2}{3} \\ \Rightarrow 8^{{\log }_8\left(\frac{\mathrm{m}}{6}\right)}=8^{\frac{2}{3}}=4 \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that}, {\mathrm{a}}^{{\log }_{\mathrm{a}}\mathrm{n}}=\mathrm{n},\text{ where }\mathrm{a}\ne 1 \\ \Rightarrow \frac{\mathrm{m}}{6}=4\Rightarrow \mathrm{m}=24 \end{gathered}$$