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If $\log _{8} \mathrm{~m}+\log _{8} \frac{1}{6}=\frac{2}{3}$, then $\mathrm{m}$ is equal to
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24
$\log _{8} \mathrm{~m}+\log _{8} \frac{1}{6}=\frac{2}{3}$
$\Rightarrow \quad \log _{8}\left(\mathrm{~m}, \frac{1}{6}\right)=\frac{2}{3}$
$\Rightarrow(8)^{\frac{2}{3}}=\mathrm{m} \cdot \frac{1}{6}$
$\Rightarrow \mathrm{m}=24$
$\Rightarrow \quad \log _{8}\left(\mathrm{~m}, \frac{1}{6}\right)=\frac{2}{3}$
$\Rightarrow(8)^{\frac{2}{3}}=\mathrm{m} \cdot \frac{1}{6}$
$\Rightarrow \mathrm{m}=24$
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