已知loga1/√2-1已知loga1/√2-1已知loga1/√2-11/(√2-1)=√2+1>1所以lg[1/(√2-1)]>0换底公式lg[1/(√2-1)]/lga所以1/lga<1/lgb<0都是负数所以lga<0,lga<0则lgalgb>0用乘以lgalgblgblgx是增函数所以b
已知loga1/√2-11/(√2-1)=√2+1>1所以lg[1/(√2-1)]>0换底公式lg[1/(√2-1)]/lga所以1/lga<1/lgb<0都是负数所以lga<0,lga<0则lgalgb>0用乘以lgalgblgblgx是增函数所以b