0,\; (t.k.\; 3>2,a\; log_22=1)\\\\log_{12}{18}=\frac{log_2{18}}{log_2{12}}=\frac{log_2{(3^2\cdot 2)}}{log_2{(2^2\cdot 3)}}=\frac{1+2log_23}{2+log_23}=\frac{1+2t}{2+t},\; t=log_23>0\\\\log_{24}{72}-log_{12}{18}=\frac{3+2t}{3+t}-\frac{1+2t}{2+t}=\frac{3}{(3+t)(2+t)}>0\; \; \to \; \; \\\\log_{24}{72}>log_{12}{18}" alt="=\frac{3+2log_23}{3+log_23}=\frac{3+2t}{2+t},\; t=log_23>0,\; (t.k.\; 3>2,a\; log_22=1)\\\\log_{12}{18}=\frac{log_2{18}}{log_2{12}}=\frac{log_2{(3^2\cdot 2)}}{log_2{(2^2\cdot 3)}}=\frac{1+2log_23}{2+log_23}=\frac{1+2t}{2+t},\; t=log_23>0\\\\log_{24}{72}-log_{12}{18}=\frac{3+2t}{3+t}-\frac{1+2t}{2+t}=\frac{3}{(3+t)(2+t)}>0\; \; \to \; \; \\\\log_{24}{72}>log_{12}{18}" align="absmiddle" class="latex-formula">