把原式用公式展开
(lga+lgx)(lgb+lgx)+1=0
(lgx)^2+(lga+lgb)lgx+lgalgb+1=0
若lg(ax)lg(bx)+1=0有解
则必存在lgx满足①式
Δ≥0
(lga+lgb)^2-4(lgalgb+1)≥0
(lga-lgb)^2≥4
[lg(a/b)]^2≥4
lg(a/b)≥2 ,lg(10^2)=2,所以a/b≥100或 lg(a/b)≤-2 lg(1/100)=-2,所以
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