高中數(shù)學(xué)對(duì)數(shù)函數(shù)—對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用蘇教版必修一

1、用心愛(ài)心專(zhuān)心對(duì)數(shù)函數(shù)對(duì)數(shù)函數(shù)—對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用一課題：對(duì)數(shù)函數(shù)——對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用二教學(xué)目標(biāo)：1.復(fù)習(xí)鞏固對(duì)數(shù)函數(shù)的圖象和性質(zhì)；2.會(huì)利用對(duì)數(shù)函數(shù)的性質(zhì)（單調(diào)性）比較兩個(gè)對(duì)數(shù)值的大小。三教學(xué)重、難點(diǎn)：對(duì)數(shù)函數(shù)性質(zhì)的靈活運(yùn)用。四教學(xué)過(guò)程：（一）復(fù)習(xí)：1對(duì)數(shù)函數(shù)的概念；2根據(jù)對(duì)數(shù)函數(shù)的圖象，敘述對(duì)數(shù)函數(shù)的性質(zhì)。（二）新課講解：例1比較下列各組數(shù)中兩個(gè)值的大小：（1）2log3.4，2log8.5；（2）0.3log1.

2、8，0.3log2.7；（3）log5.1a，log5.9a.解：（1）對(duì)數(shù)函數(shù)2logyx?在(0)??上是增函數(shù)，于是2log3.4?2log8.5；（2）對(duì)數(shù)函數(shù)0.3logyx?在(0)??上是減函數(shù)，于是0.3log1.8?0.3log2.7；（3）當(dāng)1a?時(shí)，對(duì)數(shù)函數(shù)logayx?在(0)??上是增函數(shù)，于是log5.1a?log5.9a，當(dāng)1oa??時(shí)，對(duì)數(shù)函數(shù)logayx?在(0)??上是減函數(shù)，于是log5.1a?lo

3、g5.9a說(shuō)明：本例是利用對(duì)數(shù)函數(shù)的增減性比較兩個(gè)對(duì)數(shù)的大小的，底數(shù)與1的大小關(guān)系不明確時(shí)，要分情況對(duì)底數(shù)進(jìn)行討論來(lái)比較兩個(gè)對(duì)數(shù)的大小。例2比較下列比較下列各組數(shù)中兩個(gè)值的大?。海?）6log7，7log6；（2）3log?，2log0.8；用心愛(ài)心專(zhuān)心∴44loglognm?，∴1mn??當(dāng)01m??，01n??時(shí)，得44110loglogmn??，∴44loglognm?，∴01nm???當(dāng)01m??，1n?時(shí)，得4log0m?，4

4、0logn?，∴01m??，1n?，∴01mn???綜上所述，m，n的大小關(guān)系為1mn??或01nm???或01mn???五課堂練習(xí)：1已知11loglog033ab??，則下列不等式成立的是（A）()A01ba???()B01ab???()C1ba??()D1ab??2已知01a??，1b?，1ab?，則下列不等式成立的是（B）()A11logloglogbaabbb??()B11logloglogababbb??()C11loglo