解:∵f(x)=|lgx|.∴当f(a)=f(b)时谨拿核,lga=lgb或lga=-lgb又∵0<b<a ∴lga=-lgb∴lga+lgb=lg(ab)=0即ab=1∵a+2b≥2 a•2b=2√2∴敏旁a+2b的取值范围是[2√2,+∞)故答案为:[2√2,+∞祥掘)