log2 x- log4 9 + 1/log3 2 + 2 =6
Пожалуйста! !!!!!!!

формулы :loga b -loga c=loga b/cloga b+loga c=loga b*c1/logb a=loga bloga^n (b^n) =loga bРешение :log2 x-log2 3 +log2 3 +2=6log2 ((x/3)*3)=6-2log2 x =4x=16

6=log2 64 2=log2 4 log2 x- log2 3 +log2 3 +log2 4= log2 64 log2 x*4 =log2 64 x*4=64 x=64/4=16