Trong mỗi trường hợp sau, hãy tính \({\log _a}x\) biết \({\log _a}b = 3,{\log _a}c = - 2\):
a) \(x = {a^3}{b^2}\sqrt c ;\)
b) \(x = {{{a^4}\root 3 \of b } \over {{c^3}}}.\)
LG a
\(x = {a^3}{b^2}\sqrt c\)
Lời giải chi tiết:
\({\log _a}x = {\log _a}\left( {{a^3}{b^2}\sqrt c } \right)\)
\(\begin{array}{l}
= {\log _a}{a^3} + {\log _a}{b^2} + {\log _a}\sqrt c \\
= 3{\log _a}3 + 2{\log _a}b + {\log _a}{c^{\frac{1}{2}}}
\end{array}\)
\(= 3 + 2{\log _a}b + {1 \over 2}{\log _a}c \)
\(= 3 + 2.3 + {1 \over 2}\left( { - 2} \right) = 8\).
LG b
\(x = {{{a^4}\root 3 \of b } \over {{c^3}}}.\)
Lời giải chi tiết:
\({\log _a}x = {\log _a}\left( {{{{a^4}\root 3 \of b } \over {{c^3}}}} \right)\)
\(\begin{array}{l}
= {\log _a}\left( {{a^4}\sqrt[3]{b}} \right) - {\log _a}{c^3}\\
= {\log _a}{a^4} + {\log _a}\sqrt[3]{b} - {\log _a}{c^3}\\
= 4{\log _a}a + {\log _a}{b^{\frac{1}{3}}} - {\log _a}{c^3}
\end{array}\)
\( = 4 + {1 \over 3}{\log _a}b - 3{\log _a}c \)
\(= 4 + {1 \over 3}.3 - 3\left( { - 2} \right) = 11\).
