Cho logab = 5; logac = 9 (b, c > 0; 0 < a ≠ 1). Khi đó a) logabc = −4.
Giải thích
a) logabc = logab + logac = 5 + 9 = 14.
b) \({\log _b}c = \frac{{{{\log }_a}c}}{{{{\log }_a}b}} = \frac{9}{5}\).
c) \({\log _a}\left( {\frac{b}{{{c^2}}}} \right) = {\log _a}b - {\log _a}{c^2} = {\log _a}b - 2{\log _a}c = 5 - 2.9 = - 13\).
d) \({\log _{{a^3}}}\frac{{\sqrt b }}{{{c^2}}} = {\log _{{a^3}}}\sqrt b - {\log _{{a^3}}}{c^2}\)\( = \frac{1}{6}{\log _a}b - \frac{2}{3}{\log _a}c\)\( = \frac{5}{6} - \frac{{18}}{3} = - \frac{{31}}{6}\).
Đáp án: a) Sai; b) Sai; c) Sai; d) Đúng.