Đáp án:
$\begin{array}{l}
6) + ){\log _9}6 = \frac{{{{\log }_3}6}}{{{{\log }_3}9}} = \frac{{1 + {{\log }_3}2}}{2} = a\\
\Rightarrow {\log _3}2 = 2a - 1\\
\Rightarrow {\log _2}3 = \frac{1}{{2a - 1}}\\
+ ){\log _{18}}32 = \frac{{{{\log }_2}32}}{{{{\log }_2}18}}\\
= \frac{5}{{{{\log }_2}\left( {{{2.3}^2}} \right)}} = \frac{5}{{1 + 2{{\log }_2}3}}\\
= \frac{5}{{1 + \frac{2}{{2a - 1}}}} = \frac{{5\left( {2a - 1} \right)}}{{2a + 1}}\\
= \frac{{10a - 5}}{{2a + 1}}\\
7)\\
\log 56 = \log 7 + \log 8 = \log 2.{\log _2}7 + 3\log 2 = a.b + 3a
\end{array}$
