Given that a, B and X are positive numbers, LG (AX) LG (BX) + 1 = 0, the range of a / B is obtained? LG (BX) LG (AX) + 1 = 0, and a, B, X are positive numbers Then (LGA + lgx) (LGB + lgx) + 1 = 0 (lgx)^2+（lga+lgb）lgx+1+lgalgb=0 This equation has a solution So (LGA + LGB) ^ 2-4lgalgb-4 ≥ 0 (lga)^2+2lgalhb+(lgb)^2-4lgalgb-4≥0 (lga-lgb)^2≥4 LGA LGB ≥ 2 or LGA LGB ≤ - 2 LG (a-b) ≥ 2 or LGA / b ≤ - 2 So a / b ≥ 100 or 0

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