Answer
$\dfrac{2}{3}$
Work Step by Step
RECALL:
(1) $\log_a{P} - \log_a{Q} = \log_a{(\frac{P}{Q})}$
(2) $\log_a{P} + \log_a{Q} = \log_a{(PQ)}$
Use rule (1) above to obtain:
$=\log_8{(\frac{6}{3})} + \log_8{2}
\\=\log_8{2} + \log_8{2}$
Use rule (2) above to obtain:
$=\log_8{(2\cdot2)}
\\=\log_8{4}$
RECALL:
$\log_a{b} = \dfrac{\log{b}}{\log{a}}$
Use the rule above to obtain:
$\log_8{4} = \dfrac{\log{4}}{\log{8}}$
Since $4=2^2$ and $8=2^3$, then the expression above is equivalent to:
$=\dfrac{\log{(2^2)}}{\log{(2^3)}}$
RECALL:
$\log{(M^r)}=r \cdot \log{M}$
Use the rule above to obtain:
$\require{cancel}
\dfrac{\log{(2^2)}}{\log{(2^3)}}
\\=\dfrac{2\log{2}}{3\log{2}}
\\=\dfrac{2\cancel{\log{2}}}{3\cancel{\log{2}}}
\\=\dfrac{2}{3}$
