Giải bài 6 trang 90 – SGK môn Giải tích lớp 12
Cho \({{\log }_{a}}b=3,\,{{\log }_{a}}c=-2\). Hãy tính \({{\log }_{a}}x\) với:
a) \(x={{a}^{3}}{{b}^{2}}\sqrt{c};\)
b) \(x=\dfrac{{{a}^{4}}\sqrt[3]{b}}{{{c}^{3}}}.\)
a) Ta có:
\(\begin{aligned} &{{\log }_{a}}\left( {{a}^{3}}{{b}^{2}}\sqrt{c} \right)={{\log }_{a}}{{a}^{3}}+{{\log }_{a}}{{b}^{2}}+{{\log }_{a}}\sqrt{c} \\ & =3{{\log }_{a}}a+2{{\log }_{a}}b+\dfrac{1}{2}{{\log }_{a}}c \\ & =3+2.3+\dfrac{1}{2}.\left( -2 \right)=8 \\ \end{aligned} \)
b) Ta có:
\(\begin{aligned} & {{\log }_{a}}\left( \dfrac{{{a}^{4}}\sqrt[3]{b}}{{{c}^{3}}} \right)={{\log }_{a}}{{a}^{4}}+{{\log }_{a}}\sqrt[3]{b}-{{\log }_{a}}{{c}^{3}} \\ & =4{{\log }_{a}}a+\dfrac{1}{3}{{\log }_{a}}b-3{{\log }_{a}}c \\ & =4+\dfrac{1}{3}.3-3.\left( -2 \right)=11 \\ \end{aligned} \)
Ghi nhớ: Giả sử \(a,b_1,b_2,b\) là các số dương và \(a\ne 1\) thì
\({{\log }_{a}}\left( {{b}_{1}}{{b}_{2}} \right)={{\log }_{a}}{{b}_{1}}+{{\log }_{a}}{{b}_{2}}; \,\,\,\,\, {{\log }_{a}}\left( \dfrac{{{b}_{1}}}{{{b}_{2}}} \right)={{\log }_{a}}{{b}_{1}}-{{\log }_{a}}{{b}_{2}};\,\,\,\,\log_ab^\alpha=\alpha\log_ab.\)