Đáp án:
$\begin{array}{l}
8)\\
+ ){\log _{12}}18 = b \Rightarrow \frac{{{{\log }_2}18}}{{{{\log }_2}12}} = b = \frac{{1 + 2{{\log }_2}3}}{{2 + {{\log }_2}3}}\\
\Rightarrow {\log _2}3 = \frac{{2b - 1}}{{2 - b}}\\
+ ){\log _6}15 = \frac{{{{\log }_2}15}}{{{{\log }_2}6}} = \frac{{{{\log }_2}3 + {{\log }_2}5}}{{{{\log }_2}3 + 1}} = a\\
\Rightarrow {\log _2}5 = \left( {a - 1} \right){\log _2}3 + a\\
= \left( {a - 1} \right).\frac{{2b - 1}}{{2 - b}} + a = \frac{{a\left( {b + 1} \right) - 2b + 1}}{{2 - b}}\\
+ ){\log _{25}}24 = \frac{{{{\log }_2}24}}{{{{\log }_2}25}} = \frac{{3 + {{\log }_2}3}}{{2{{\log }_2}5}}\\
= \frac{{3 + \frac{{2b - 1}}{{2 - b}}}}{{2.\frac{{a\left( {b + 1} \right) - 2b + 1}}{{2 - b}}}} = \frac{{5 - b}}{{2a.\left( {b + 1} \right) - 4b + 2}}\\
9)\\
+ )\log 5 = a \Rightarrow \log \frac{{10}}{2} = a\\
\Rightarrow 1 - \log 2 = a\\
\Rightarrow \log 2 = 1 - a\\
\Rightarrow {\log _2}5 = \frac{{\log 5}}{{\log 2}} = \frac{a}{{1 - a}}\\
{\log _2}3 = \frac{{\log 3}}{{\log 2}} = \frac{b}{{1 - a}}\\
+ ){\log _{30}}8 = \frac{{{{\log }_2}8}}{{{{\log }_2}30}} = \frac{3}{{1 + {{\log }_2}3 + {{\log }_2}5}}\\
= \frac{3}{{1 + \frac{b}{{1 - a}} + \frac{a}{{1 - a}}}}\\
= \frac{{3\left( {1 - a} \right)}}{{1 - a + b + a}}\\
= \frac{{3\left( {1 - a} \right)}}{{1 + b}}
\end{array}$
