Cho \(a = {\log _2}5,b = {\log _3}5\). Tính \({\lo...

D

- Hướng dẫn giải

Ta có \({\log _{24}}600 = \frac{{{{\log }_5}600}}{{{{\log }_5}24}} = \frac{{{{\log }_5}{5^2}.24}}{{{{\log }_5}24}} = \frac{{2 + {{\log }_5}24}}{{{{\log }_5}24}}\)

Mà \({\log _5}24 = {\log _5}{2^3}.3 = 3{\log _5}2 + {\log _5}3 = \frac{3}{a} + \frac{1}{b} = \frac{{a + 3b}}{{ab}}\)

Do đó \({\log _{24}}600 = \frac{{2 + \frac{{a + 3b}}{{ab}}}}{{\frac{{a + 3b}}{{ab}}}} \Rightarrow {\log _{24}}600 = \frac{{2ab + a + 3b}}{{a + 3b}}.\)