logbc的精選

問題詳情：若a＞b＞0，0＜c＜1，則（）A．logac＜logbc                                             B．logca＜logcb C．ac＜bc                               ...

問題詳情：設1＜a≤b≤c，*：logab＋logbc＋logca≤logba＋logcb＋logac.【回答】 (2)令logab＝x，logbc＝y，則logca＝，logba＝，logcb＝，logac＝xy.由1＜a≤b≤c得x＝logab≥1，y＝logbc≥1.由(1)亦即得到所*的不等式成立．知識點：不等式題型：解答題...