Question

Помогите пожалуйста ) $(1-log _{2} 12)(1-log_{6}12 )$

Answer 1

(1-log 2 12)(1- log 6 12)= (1-log 2 (2*6))(1-log 6 (2*6))=(1-(log 2 2+log 2 6))(1-(log 6 2+log 6 6))=(1-1-log 2 6)(1-log 6 2-1)=log 2 6*log 6 2=log 2 6* 1/log 2 6=1