Tính:
a) \({\log _3}5.{\log _5}7.{\log _7}9\);
b) \({\log _2}\frac{1}{{25}}.{\log _3}\frac{1}{{32}}.{\log _5}\frac{1}{{27}}\).
Sử dụng kiến thức về phép tính lôgarit để tính: Cho các số dương a, b, N, \(a \ne 1,b \ne 1\) ta có: \({\log _a}N = \frac{{{{\log }_b}N}}{{{{\log }_b}a}}\).
a) \({\log _3}5.{\log _5}7.{\log _7}9\)
\( = {\log _3}5.\frac{{{{\log }_3}7}}{{{{\log }_3}5}}.\frac{{{{\log }_3}9}}{{{{\log }_3}7}} \\ = {\log _3}9 \\ = {\log _3}{3^2} \\ = 2;\)
b) \({\log _2}\frac{1}{{25}}.{\log _3}\frac{1}{{32}}.{\log _5}\frac{1}{{27}} \)
\( = {\log _2}{5^{ - 2}}.{\log _3}{2^{ - 5}}.{\log _5}{3^{ - 3}} \\ = - 2{\log _2}5.\left( { - 5} \right){\log _3}2.\left( { - 3} \right){\log _5}3\\ = - 30{\log _2}5.{\log _3}2.{\log _5}3 \\ = - 30.{\log _2}5.\frac{{{{\log }_2}2}}{{{{\log }_2}3}}.\frac{{{{\log }_2}3}}{{{{\log }_2}5}} \\ = - 30\)