Cho \(a = {\log _3}2,b = {\log _3}5.\) Khi đó \(\log 60\) bằng

* Hướng dẫn giải

\(\log 60 = \frac{{{{\log }_3}60}}{{{{\log }_3}10}} = \frac{{{{\log }_3}{2^2} + {{\log }_3}3 + {{\log }_3}5}}{{{{\log }_3}2 + {{\log }_3}5}} = \frac{{2{{\log }_3}2 + 1 + {{\log }_3}5}}{{{{\log }_3}2 + {{\log }_3}5}} = \frac{{2a + b + 1}}{{a + b}}\)