Toán 12 Tính \({\log_6}90\) theo a, b

Cho \(\log 2 = a;log3 = b.\) Tính \({\log_6}90\) theo a, b.
A. \(lo{g_6}90 = \frac{{2b - 1}}{{a + b}}\)
B. \(lo{g_6}90 = \frac{{b+1}}{{a + b}}\)
C. \(lo{g_6}90 = \frac{{2b +1}}{{a + b}}\)
D. \(lo{g_6}90 = \frac{{2b + 1}}{{a +2 b}}\)

\({\log _6}90 = \frac{{\log 90}}{{\log 6}} = \frac{{\log 9 + \log 10}}{{\log 2 + \log 3}} = \frac{{2b + 1}}{{a + b}}\)