Hãy tính \({\log _{140}}63\) theo a, b, c

Hàm số mũ | Hàm số lũy thừa | Hàm số mũ và lũy thừa | hàm số loagrit | logarit |
Cho \(a = {\log _2}3;\;\,\,b = {\log _3}5;\;\,\,c = {\log _7}2.\) Hãy tính \({\log _{140}}63\) theo a, b, c.
A. \({\log _{140}}63 = \frac{{2ac + 1}}{{abc + 2c + 1}}.\)
B. \({\log _{140}}63 = \frac{{2ac + 1}}{{abc + 2c -1}}.\)
C. \({\log _{140}}63 = \frac{{2ac + 1}}{{abc - 2c + 1}}.\)
D. \({\log _{140}}63 = \frac{{2ac - 1}}{{abc + 2c + 1}}.\)

Ta có: \({\log _{140}}63 = {\log _{140}}9 + {\log _{140}}7 = 2{\log _{140}}9 + {\log _{140}}7\)
\(2{\log _{140}}3 = \frac{2}{{{{\log }_3}140}} = \frac{2}{{{{\log }_3}7 + 2{{\log }_3}2 + {{\log }_3}5}}\)
\(= \frac{2}{{{{\log }_3}2.{{\log }_2}7 + \frac{2}{a} + b}} = \frac{2}{{\frac{1}{{ca}} + \frac{2}{a} + b}} = \frac{{2ca}}{{1 + 2c + abc}}\)
\({\log _{140}}7 = \frac{1}{{{{\log }_7}140}} = \frac{1}{{{{\log }_7}7 + 2{{\log }_7}2 + {{\log }_7}5}}\)
\(= \frac{1}{{1 + 2c + {{\log }_7}2.{{\log }_2}3.{{\log }_3}5}} = \frac{1}{{1 + 2c + abc}}\)
Vậy: \({\log _{140}}63 = \frac{{2ca}}{{1 + 2c + abc}} + \frac{1}{{1 + 2c + abc}} = \frac{{2ac + 1}}{{1 + 2c + abc}}.\)