lg(bx)lg(ax)+1=0,且a,b,x為正數(shù)
則(lga+lgx)(lgb+lgx)+1=0
(lgx)^2+（lga+lgb）lgx+1+lgalgb=0
這個方程有解
所以（lga+lgb）^2-4lgalgb-4≥0
(lga)^2+2lgalhb+(lgb)^2-4lgalgb-4≥0
(lga-lgb)^2≥4
lga-lgb≥2或 lga-lgb≤-2
lg(a-b)≥2或 lga/b≤-2
所以a/b≥100 或0