已知a,b,x为正数,且lg(bx).lg(ax)+1=0,求a/b取值范围

lg(bx)lg(ax)+1=0,且a,b,x为正数
则(lga+lgx)(lgb+lgx)+1=0
(lgx)^2+（lga+lgb）lgx+1+lgalgb=0
这个方程有解
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<a/b≤1/100

设t=lgax,
lgbx = -1/t

ax= 10^t
bx = 10^(-1/t)
a/b = 10^t / 10^(-1/t)
= 10^(t+1/t)

t>0, t+1/t >= 2, a/b >= 100
t<0, t+1/t <= -2, a/b <= 1/100
a/b的取值范围是 (0, 1/100] , [100, +∞)

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