1<a<blga>lgb 得a>b矛盾 0<a<1<b-lga>lgb 得1/a>b ab<1 0<a<b<1得ab<1所以ab<1
f(a) = |lga|f(b) = |lgb|因为f(a)>f(b), 0<a<b所以a<1, b<1/a所以ab<1