Bài 6 trang 90 SGK Giải tích 12

Cho \({\log _a}b = 3,{\log _a}c =  - 2\) . Hãy tính \(\log_ax\) với:

a) \(x = {a^3}{b^2}\sqrt c \)

b) \(x = {{{a^4}\root 3 \of b } \over {{c^3}}}\)

\(\begin{array}{l}a)\,{\log _a}x = {\log _a}\left( {{a^3}{b^2}\sqrt c } \right)\\= {\log _a}{a^3} + {\log _a}{b^2} + {\log _a}\sqrt c \\= {\log _a}{a^3} + {\log _a}{b^2} + {\log _a}{c^{\frac{1}{2}}}\\= 3{\log _a}a + 2{\log _a}b + \dfrac{1}{2}{\log _a}c\\= 3 + 2.3 + \dfrac{1}{2}\left( { - 2} \right) = 8\\b)\,{\log _a}x = {\log _a}\dfrac{{{a^4}\sqrt[3]{b}}}{{{c^3}}}\\= {\log _a}{a^4} + {\log _a}\sqrt[3]{b} - {\log _a}{c^3}\\= {\log _a}{a^4} + {\log _a}{b^{\frac{1}{3}}} - {\log _a}{c^3}\\= 4{\log _a}a + \dfrac{1}{3}{\log _a}b - 3{\log _a}c\\= 4.1 + \dfrac{1}{3}.3 - 3\left( { - 2} \right)\\= 11\end{array}\)