Cả 2 cây a,b ạ giúp vd9

Ví dụ `9` 

$\begin{array}{l}
a){\log _a}\left( {{a^{\dfrac{1}{2}}}{b^5}} \right) = {\log _a}{a^{\dfrac{1}{2}}} + {\log _a}{b^5}\\
 = \dfrac{1}{2}{\log _a}a + 5{\log _a}b\\
 = \dfrac{1}{2} + 5.4 = \dfrac{1}{2} + 20 = \dfrac{{41}}{2}\\
b){\log _{{a^3}{b^2}}}\left( {{a^2}{b^3}} \right) = {\log _{{a^3}{b^2}}}{a^2} + {\log _{{a^3}{b^2}}}{b^3}\\
 = \dfrac{1}{{{{\log }_{{a^2}}}\left( {{a^3}{b^2}} \right)}} + \dfrac{1}{{{{\log }_{{b^3}}}\left( {{a^3}{b^2}} \right)}}\\
 = \dfrac{2}{{{{\log }_a}\left( {{a^3}{b^2}} \right)}} + \dfrac{3}{{{{\log }_b}\left( {{a^3}{b^2}} \right)}}\\
 = \dfrac{2}{{{{\log }_a}{a^3} + {{\log }_a}{b^2}}} + \dfrac{3}{{{{\log }_b}{a^3} + {{\log }_b}{b^2}}}\\
 = \dfrac{2}{{3 + 2{{\log }_a}b}} + \dfrac{3}{{3{{\log }_b}a + 2}}\\
 = \dfrac{{24}}{{11}}
\end{array}$