Question

Срочно! Сравните два лёгких логарифма!

Answer 1

$log_{24}{72}=frac{log_2{72}}{log_2{24}}=frac{log_{2}{(2^3cdot 9)}}{log_{2}{(2^3cdot 3)}}=$

$=frac{3+2log_23}{3+log_23}=frac{3+2t}{2+t},; t=log_23>0,; (t.k.; 3>2,a; log_22=1)\\log_{12}{18}=frac{log_2{18}}{log_2{12}}=frac{log_2{(3^2cdot 2)}}{log_2{(2^2cdot 3)}}=frac{1+2log_23}{2+log_23}=frac{1+2t}{2+t},; t=log_23>0\\log_{24}{72}-log_{12}{18}=frac{3+2t}{3+t}-frac{1+2t}{2+t}=frac{3}{(3+t)(2+t)}>0; ; to ; ; \\log_{24}{72}>log_{12}{18}$