Đáp án đúng: B
Phương pháp giải:
Sử dụng các công thức logarit:
\(\begin{array}{l}{\log _a}x + {\log _a}y = {\log _a}xy\,\,\,\left( {0 < a
e 1,\,\,x,\,\,y > 0} \right)\\{\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}\,\,\left( {0 < a,\,\,c
e 1,\,\,b > 0} \right)\\{\log _a}b = \frac{1}{{{{\log }_b}a}}\,\,\,\left( {0 < a,\,\,b
e 1} \right)\end{array}\)
Giải chi tiết:\(\begin{array}{l} + )\,\,\,a = {\log _2}60 = {\log _2}\left( {{2^2}.15} \right)\\\,\,\,\, = {\log _2}{2^2} + {\log _2}15 = 2 + {\log _2}15\\ \Rightarrow {\log _2}15 = a - 2\\ \Rightarrow {\log _2}5 = \frac{{{{\log }_{15}}5}}{{{{\log }_{15}}2}} = \frac{{\frac{1}{{{{\log }_5}15}}}}{{\frac{1}{{{{\log }_2}15}}}} = \frac{{{{\log }_2}15}}{{{{\log }_5}15}} = \frac{{a - 2}}{b}\\ + )\,\,\,b = {\log _5}15 = {\log _5}\left( {3.5} \right)\\\,\,\,\,\,\,\,\,\,\,\, = {\log _5}3 + {\log _5}5 = 1 + {\log _5}3\\ \Rightarrow {\log _5}3 = b - 1\\ \Rightarrow {\log _2}3 = {\log _2}5.{\log _5}3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{a - 2}}{b}.\left( {b - 1} \right) = \frac{{ab - 2b - a + 2}}{b}\\ \Rightarrow {\log _2}12 = {\log _2}\left( {{2^2}.3} \right) = {\log _2}{2^2} + {\log _2}3 = 2 + {\log _2}3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2b + ab - 2b - a + 2}}{b} = \frac{{ab - a + 2}}{b}.\end{array}\)
Vậy \(P = \frac{{ab - a + 2}}{b}.\)
Chọn B.
