Giải bài 35 trang 92 SGK giải tích nâng cao 12
Trong mỗi trường hợp sau đây, hãy tính \({{\log }_{a}}x\), biết \({{\log }_{a}}b=3,{{\log }_{a}}c=-2 \)
a) \(x={{a}^{3}}{{b}^{2}}\sqrt{c}\); b) \(x=\dfrac{{{a}^{4}}\sqrt[3]{b}}{{{c}^{3}}}\)
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
\(\log_{a}\left( bc \right)={\log }_{a}{b}+{\log }_{a}{c}; \,\,\,\,\,\, {\log _{a}\left( \dfrac{b}{c} \right)={\log }_{a}}{{b}}-{\log }_{a}{c} \)
\({{\log }_{a}}\left( {{b}^{\alpha }} \right)=\alpha {{\log }_{a}}b;\,\,\,\,\,\,\,{{a}^{{{\log }_{a}}b}}=b;\,\,\,\,\,\)