Đặt a=log 27 5, b=log 8 7, c=log 2 3. Khi đó log 12 35 bằng

Tacó: $a=\log _{27} 5=\log _{3^3} 5=13 \log _{3} 5 \Leftrightarrow 3 a=\log _{3} 5$
$b=\log _{8} 7=\log _{2^3} 7=13 \log _{2} 7 \Leftrightarrow 3 b=\log _{2} 7$$\mathrm{c}=\log _{2} 3$
Biến đồi: $\log _{12} 35=\log _{12} 5+\log _{12} 7=\log 35 \log 312+\log 27 \log 212=3 {blog3}(3.4)+3 {alog2}(4.3)=$
$3{a} 1+2{c}+3{~b} 2+{c}=3{ac}+3{bc}+2$