Ta có :
\(a=\log_{12}18=\frac{\log_218}{\log_212}=\frac{\log_2\left(2.3^2\right)}{\log_2\left(2^2.3\right)}=\frac{1+2\log_23}{2+\log_23}\)
\(\Rightarrow a\left(a+\log_23\right)=1+2\log_23\Leftrightarrow\log_23=\frac{1-2a}{a-2}\left(1\right)\)
\(b=\log_{24}54=\frac{\log_254}{\log_224}=\frac{\log_2\left(2.3^2\right)}{\log_2\left(2^2.3\right)}=\frac{1+3\log_23}{3+\log_23}\)
\(\Rightarrow b\left(3+\log_23\right)=1+3\log_23\Leftrightarrow\log_23=\frac{1-3b}{b-3}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{1-2a}{a-2}=\frac{1-3b}{b-3}\Leftrightarrow\left(1-2a\right)\left(b-3\right)=\left(1-3b\right)\left(a-2\right)\)
\(\Leftrightarrow ab+5\left(a-b\right)=1\Rightarrow\) Điều phải chứng minh