log2 x + log3 x< log3 6

ОДЗ:x>0log 2 (x)=(log 3 (x))/(log 3 (2)(log 3 (x))/(log 3 (2)+log 3 (x) < log 3 (6) log 3 (x) (1/log 3 (2) +1)< log 3 (6) log 3 (x) (log 3 (2) +1)/(log 3 (2)< log 3 (6)log 3 (x) (log 3 (6))/log 3 (2)<log 3 (6)|: log 3 (6)>0(log 3 (x))/(log 3 (2)) <1 |*  log 3 (2)>0log 3 (x)<log 3 (2)x<2ОДЗ: x>0Ответ: х∈(0; 2)