Toán 12 Biểu diễn \({\log _{12}}35\) theo a, b, c

Cho \({\log _{27}}5 = a,\,{\log _8}7 = b,{\log _2}3 = c\). Biểu diễn \({\log _{12}}35\) theo a, b, c.
A. \({\log _{12}}35 = \frac{{3b + 3ac}}{{c + 2}}\)
B. \({\log _{12}}35 = \frac{{3b + 2ac}}{{c + 2}}\)
C. \({\log _{12}}35 = \frac{{3b + 2ac}}{{c + 3}}\)
D. \({\log _{12}}35 = \frac{{3b + 3ac}}{{c + 1}}\)

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