Question

Здравствуйте! Столкнулся с проблемой - (log(3)2+log(2)81+4)*(log(3)2-2log(18)2)*log(2)3-log(3)2=...(число) - основание..

Answer 1

$(log_{3}2+log_{2}81+4)(log_{3}2-2log_{18}2)log_{2}3-log_{3}2=\\ (\frac{1}{log_{2}3}+log_{2}81+4)(log_{3}2-log_{18}4)*log_{2}3-log_{3}2=\\ (\frac{1}{log_{2}3}+log_{2}81+4)(\frac{1}{log_{2}3}-\frac{2}{log_{2}(9*2)})*log_{2}3-log_{3}2=\\ (\frac{1}{log_{2}3}+4log_{2}3+4)(\frac{1}{log_{2}3}-\frac{2}{2log_{2}3+1})*log_{2}3-\frac{1}{log_{2}3}=$

удобно заменить  $log_{2}3=t\\ (\frac{1}{t}+4t+4)(\frac{1}{t}-\frac{2}{2t+1})*t-\frac{1}{t} =2$
 Ответ 2