a) Ta có:
\(\displaystyle a = {\log _3}15 = {\log _3}(3.5)\)\(\displaystyle = {\log _3}3 + {\log _3}5 = 1 + {\log _3}5\) \(\displaystyle \Rightarrow {\log _3}5 = a - 1\)
Do đó:
\(\displaystyle{\log _{\sqrt 3 }}50 = {\log _{{3^{\frac{1}{2}}}}}50\)\(\displaystyle = 2{\log _3}50 = 2\left( {{{\log }_3}5 + {{\log }_3}10} \right)\)\(\displaystyle = 2{\log _3}5 + 2{\log _3}10\)\(\displaystyle = 2\left( {a - 1} \right) + 2b = 2a + 2b - 2\).
b) Ta có: \(\displaystyle{\log _{140}}63 = {\log _{140}}({3^2}.7)\)\(\displaystyle = 2{\log _{140}}3 + {\log _{140}}7\)
\(\displaystyle = \frac{2}{{{{\log }_3}140}} + \frac{1}{{{{\log }_7}140}}\)\(\displaystyle = \frac{2}{{{{\log }_3}({2^2}.5.7)}} + \frac{1}{{{{\log }_7}({2^2}.5.7)}}\)
\(\displaystyle = \frac{2}{{2{{\log }_3}2 + {{\log }_3}5 + {{\log }_3}7}}\)\(\displaystyle + \frac{1}{{2{{\log }_7}2 + {{\log }_7}5 + 1}}\)
Từ đề bài suy ra:
\(\displaystyle{\log _3}2 = \frac{1}{{{{\log }_2}3}} = \frac{1}{a}\)
\(\displaystyle{\log _7}5 = {\log _7}2.{\log _2}3.{\log _3}5 = cab\)
\(\displaystyle{\log _3}7 = \frac{1}{{{{\log }_7}3}} = \frac{1}{{{{\log }_7}2.{{\log }_2}3}} = \frac{1}{{ca}}\)
Vậy \(\displaystyle{\log _{140}}63\)\(\displaystyle = \frac{2}{{\frac{2}{a} + b + \frac{1}{{ca}}}} + \frac{1}{{2c + cab + 1}}\) \(\displaystyle = \frac{{2ac + 1}}{{abc + 2c + 1}}\).
