Tính {log _{12}}35 theo a, b, c biết {log _{27}}5 = a,{log _8}7 = b,{log _2}3 = c

Ta có: \(a = {\log _{27}}5 = {\log _{{3^3}}}5 = \frac{1}{3}{\log _3}5,b = {\log _8}7 = {\log _{{2^3}}}7 = \frac{1}{3}{\log _2}7\)

\({\log _{12}}35 = \frac{{{{\log }_2}35}}{{{{\log }_2}12}} = \frac{{{{\log }_2}7 + {{\log }_2}5}}{{{{\log }_2}({{3.2}^2})}} = \frac{{{{\log }_2}7 + {{\log }_2}3.{{\log }_3}5}}{{{{\log }_2}3 + {{\log }_2}{2^2}}} = \frac{{3b + c.3a}}{{c + 2}} = \frac{{3b + 3ac}}{{c + 2}}.\)