[m]lg18=lg2\cdot 9=lg2+2lg3[/m]
[m]lg12=lg4\cdot 3=2lg2+lg3[/m]
[m]log_{12}18=\frac{lg2+2lg3}{2lg2+lg3}[/m] ⇒ [m]\frac{lg2+2lg3}{2lg2+lg3}=x[/m] ⇒
[m]lg2+2lg3=2xlg2+xlg3[/m]
[m](x-2)lg3=(1-2x)lg2[/m]
[m]lg2=\frac{x-2}{1-2x}lg3[/m]
[m]log_{6}9=\frac{lg9}{lg6}=\frac{2lg3}{lg2+lg3}=\frac{2lg3}{\frac{x-2}{1-2x}lg3+lg3}=\frac{2}{\frac{x-2}{1-2x}+1}=\frac{2(1-2x)}{-x-1}=\frac{4x-2}{x+1}[/m]
О т в е т. С)