Cho $lo{{g}_{27}}\text{5}=a, lo{{g}_{8}}7=b, lo{{g}_{2}}3=c.~$...

Phương pháp:
Sử dụng các công thức logarit
$lo{{g}_{a}}\left( xy \right)=lo{{g}_{a}}x+lo{{g}_{a}}y,lo{{g}_{a}}\left( \dfrac{x}{y}~ \right)=lo{{g}_{a}}x-lo{{g}_{a}}y,lo{{g}_{a}}b=\dfrac{1}{lo{{g}_{b}}\text{a }},lo{{g}_{a}}b=lo{{g}_{a}}c.{{\log }_{c}}\text{b}$
(Giả sử các biểu thức là có nghĩa).
Cách giải:
Ta có $lo{{g}_{27}}5=a\Leftrightarrow \dfrac{1}{3}lo{{g}_{3}}5=a\Leftrightarrow lo{{g}_{3}}5=3a~$
$~lo{{g}_{2}}3=c\Leftrightarrow lo{{g}_{2}}5=lo{{g}_{2}}3.lo{{g}_{3}}5=3ac$
$lo{{g}_{8}}7=b\Rightarrow \dfrac{1}{3}lo{{g}_{2}}7=b\Leftrightarrow lo{{g}_{2}}7=3b$
$\Rightarrow lo{{g}_{3}}7=lo{{g}_{3}}2.lo{{g}_{2}}7=\dfrac{3b}{c}.~$
Ta có:
$~P=lo{{g}_{12}}35=~lo{{g}_{12}}~\left( 5.7 \right)$
⇔ $P=lo{{g}_{12}}5+~lo{{g}_{12}}7$
⇔ $P=\dfrac{1}{lo{{g}_{5}}12}+~\dfrac{1}{lo{{g}_{7~}}12}$
⇔ $P=\dfrac{1}{lo{{g}_{5}}\left( {{2}^{2}}.3 \right)}+~\dfrac{1}{lo{{g}_{7}}\left( {{2}^{2}}.3 \right)}$
⇔ $P=\dfrac{1}{lo{{g}_{5}}3+2lo{{g}_{5}}2}+~\dfrac{1}{lo{{g}_{7}}3+2lo{{g}_{7}}2}$
$\Leftrightarrow P=\dfrac{1}{\dfrac{1}{3a}+\dfrac{2}{3ac}}+\dfrac{1}{\dfrac{1}{3b}+\dfrac{2}{3b}}$
$\Leftrightarrow P=\dfrac{1}{\dfrac{c+2}{3ac}}+\dfrac{1}{\dfrac{c+2}{3b}}$
⇔ $P=\dfrac{3ac+3b}{c+2}=\dfrac{3\left( b+ac \right)}{c+2}$