Biếu diễn log_12(35) theo a b c biết log_27(5)=a log_8(7)=b log_2(3)=c

Ta có:

\({\log _{12}}35 = \frac{{{{\log }_2}35}}{{{{\log }_2}12}} = \frac{{{{\log }_2}7 + {{\log }_2}5}}{{{{\log }_2}3 + {{\log }_2}4}}\)

Mặt khác:

\({\log _{27}}5 = a \Leftrightarrow \frac{1}{3}{\log _3}5 = a \Leftrightarrow {\log _3}5 = 3a \Leftrightarrow \frac{{{{\log }_2}5}}{{{{\log }_2}3}} = 3a \Rightarrow {\log _2}5 = 3ac.\)

\({\log _8}7 = b \Leftrightarrow \frac{1}{3}{\log _2}7 = b \Rightarrow {\log _2}7 = 3b\).

Suy ra:

\({\log _{12}}35 = \frac{{3b + 3ac}}{{c + 2}}\).