[tex]3log_{\frac{1}{12}}2+log_{\frac{1}{12}}18=log_{\frac{1}{12}}2^3+log_{\frac{1}{12}}18=log_{\frac{1}{12}}8+log_{\frac{1}{12}}18=\\\\log_{\frac{1}{12}}(8\cdot18)=log_{\frac{1}{12}}144=log_{\frac{1}{12}}12^2=log_{\frac{1}{12}}(\frac{1}{12}})^{-2}=-2log_{\frac{1}{12}}\frac{1}{12}=-2[/tex]
