因为lg(ax)*lg(bx)+1=0,所以[lga+lgx]*[lgb+lgx]+1=0,所以 [lgx]^2+(lga+lgb)lgx+(lga*lgb+1)=0判别式=(lga+lgb)^2-4(lga*lgb+1)>=0,展开并配方得:(lga-lgb)^2-4>=0,所以[lg(a/b)]^2>=4所以lg(a/b)>=2或lg(a/b)=100...