Giải bài 32 trang 92 SGK giải tích nâng cao 12

$\begin{align} & a)\,{{\log }_{8}}12-{{\log }_{8}}15+{{\log }_{8}}20={{\log }_{8}}\dfrac{12.20}{15} \\ & ={{\log }_{8}}16={{\log }_{{{2}^{3}}}}{{2}^{4}}=\dfrac{4}{3} \\ & b)\,\dfrac{1}{2}{{\log }_{7}}36-{{\log }_{7}}14-3{{\log }_{7}}\sqrt[3]{21}={{\log }_{7}}6-{{\log }_{7}}14-{{\log }_{7}}21 \\ & ={{\log }_{7}}\dfrac{6}{14.21}={{\log }_{7}}\dfrac{1}{49}={{\log }_{7}}{{7}^{-2}}=-2 \\ & c)\,\dfrac{{{\log }_{5}}36-{{\log }_{5}}12}{{{\log }_{5}}9}=\dfrac{{{\log }_{5}}\dfrac{36}{12}}{{{\log }_{5}}{{3}^{2}}}=\dfrac{{{\log }_{5}}3}{2{{\log }_{5}}3}=\dfrac{1}{2} \\ &d)\, {{36}^{{{\log }_{6}}5}}+{{10}^{1-\log 2}}-{{8}^{{{\log }_{2}}3}}={{5}^{{{\log }_{6}}36}}+\dfrac{10}{{{10}^{\log 2}}}-{{3}^{{{\log }_{2}}8}} \\ & ={{5}^{2}}+\dfrac{10}{{{2}^{\log 10}}}-{{3}^{3}}=25+5-27=3 \\ \end{align}$

Ghi nhớ: Giả sử a, b, c > 0 và $a\ne 1$. Ta có

$\log_{a}\left( bc \right)={\log }_{a}{b}+{\log }_{a}{c}; \,\,\,\,\,\, {\log _{a}\left( \dfrac{b}{c} \right)={\log }_{a}}{{b}}-{\log }_{a}{c}$

${{\log }_{a}}\left( {{b}^{\alpha }} \right)=\alpha {{\log }_{a}}b;\,\,\,\,\,\,\,{{a}^{{{\log }_{a}}b}}=b;\,\,\,\,\,$