Giải bài 35 trang 92 SGK giải tích nâng cao 12

a) Với $x={{a}^{3}}{{b}^{2}}\sqrt{c}$

$\begin{align} \Rightarrow {{\log }_{a}}x &={{\log }_{a}}\left( {{a}^{3}}{{b}^{2}}\sqrt{c} \right) \\ & ={{\log }_{a}}{{a}^{3}}+{{\log }_{a}}{{b}^{2}}+{{\log }_{a}}\sqrt{c} \\ & =3+2{{\log }_{a}}b+\dfrac{1}{2}{{\log }_{a}}c \\ & =3+2.3+\dfrac{1}{2}\left( -2 \right) \\ & =8 \end{align}$

b) Với $x=\dfrac{{{a}^{4}}\sqrt[3]{b}}{{{c}^{3}}}$

$\begin{align} \Rightarrow {{\log }_{a}}x&={{\log }_{a}}\left( \dfrac{{{a}^{4}}\sqrt[3]{b}}{{{c}^{3}}} \right) \\ & ={{\log }_{a}}{{a}^{4}}+{{\log }_{a}}\sqrt[3]{b}-{{\log }_{a}}{{c}^{3}} \\ & =4+\dfrac{1}{3}{{\log }_{a}}b-3{{\log }_{a}}c \\ & =4+\dfrac{1}{3}.3-3.\left( -2 \right) \\ & =4+1+4=11 \end{align}$

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

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

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