Đặt $a = {\log _{27}}5,\,\,b = {\log _8}7,\,\,c = {\log _2}3$. Khi đó ${\log _{12}}35$ bằng:

Ta có:

$\begin{array}{l}\,\,\,{\log _{12}}35 = {\log _{12}}\left( {5.7} \right)\\ = {\log _{12}}5 + {\log _{12}}7\\ = \dfrac{1}{{{{\log }_5}\left( {{2^2}.3} \right)}} + \dfrac{1}{{{{\log }_7}\left( {{2^2}.3} \right)}}\\ = \dfrac{1}{{2{{\log }_5}2 + {{\log }_5}3}} + \dfrac{1}{{2{{\log }_7}2 + {{\log }_7}3}}\end{array}$

Theo bài ra ta có:

$\begin{array}{l}a = {\log _{27}}5 = \dfrac{1}{{{{\log }_5}{3^3}}} = \dfrac{1}{{3{{\log }_5}3}}\\ \Rightarrow {\log _5}3 = \dfrac{1}{{3a}}\\b = {\log _8}7 = \dfrac{1}{{{{\log }_7}{2^3}}} = \dfrac{1}{{3{{\log }_7}2}}\\ \Rightarrow {\log _7}2 = \dfrac{1}{{3b}}\\c = {\log _2}3 \Rightarrow {\log _3}2 = \dfrac{1}{c}\\ \Rightarrow {\log _5}2 = {\log _5}3.{\log _3}2 = \dfrac{1}{{3a}}.\dfrac{1}{c} = \dfrac{1}{{3ac}}\\\,\,\,\,\,\,{\log _7}3 = {\log _7}2.{\log _2}3 = \dfrac{1}{{3b}}.c = \dfrac{c}{{3b}}\end{array}$

Khi đó ta có:

$\begin{array}{l}\,\,\,{\log _{12}}35 = {\log _{12}}\left( {5.7} \right)\end{array}$

$= \dfrac{1}{{\dfrac{2}{{3ac}} + \dfrac{1}{{3a}}}} + \dfrac{1}{{\dfrac{2}{{3b}} + \dfrac{c}{{3b}}}}$

$= \dfrac{{3ac}}{{2 + c}} + \dfrac{{3b}}{{2 + c}} = \dfrac{{3ac + 3b}}{{c + 2}}$