Biết {log_27}5=a, {log_8}7=b, {log_2}3=c biểu diễn {log_12}35 theo a b c

Ta có: ${\log _{27}}5 = \frac{1}{3}{\log _3}5 = a \Leftrightarrow {\log _3}5 = 3a$, ${\log _8}7 = \frac{1}{3}{\log _2}7 = b \Leftrightarrow {\log _2}7 = 3b.$ 

Mà:

$\begin{array}{l} {\log _{12}}35 = \frac{{{{\log }_2}\left( {7.5} \right)}}{{{{\log }_2}\left( {{{3.2}^2}} \right)}} = \frac{{{{\log }_2}7 + {{\log }_2}5}}{{{{\log }_2}3 + 2}} = \frac{{{{\log }_2}7 + {{\log }_2}3.{{\log }_3}5}}{{{{\log }_2}3 + 2}}\\ = \frac{{3b + c.3a}}{{c + 2}} = \frac{{3\left( {b + ac} \right)}}{{c + 2}}. \end{array}$