指數(shù)與對數(shù)函數(shù).ppt

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1、對數(shù)與對數(shù)函數(shù)對數(shù)的產(chǎn)生指數(shù)式與對數(shù)式的等價轉(zhuǎn)換：ab?Nb?logaN(a?0且a?1)(a?0且a?1)對數(shù)的運算法則：loga(MN)?logaM?logaNMloga()?logaM?logaNNnNlogaM?nlogaMaloga?NNlogaa?1logaa?Nlg2?lg5?1loga1?0例1、求值：11(1)log??3273(2)log33?33(3)2log10?log0.25?255222(4)lg5?lg8?lg5?lg20?(lg2)?3334?log?48(5)2229?log2?33(6)2781(7)lg(3?5?3?5)?22x=loga

2、y互為反函數(shù)（一）對數(shù)函數(shù)的概念：2、理解：3對數(shù)函數(shù)和指數(shù)函數(shù)互為反函數(shù)結(jié)論1：定義域，值域互換y=axy=logax函數(shù)(a>0,a≠1)(a>0,a≠1)定義域(-?,+?)(0,+?)值域(0,+?)(-?,+?)預習作業(yè)：1.在同一坐標系中，畫出y=2x和它反4函數(shù)y=log2x的圖象.問題1:你是用什么方法畫出對數(shù)函數(shù)圖象的？問題2：對數(shù)函數(shù)的圖象有幾種情況？5(a＞1)(0＜a＜1)xyy=axy?ayy=xy=xy=logax1·1·O·1xO1·xy=logax（二）對數(shù)函數(shù)的圖象和性質(zhì)：6a＞10＜a＜1yy=logaxy(a＞1)圖象O·xO1·x1y=

3、logax(0＜a＜1)定義域（0，＋∞）性值域R函數(shù)同當x=1時,logax=0當x=1時,logax=0值變正當0＜x＜1時,logax＜0當0＜x＜1時,logax＞0化規(guī)異律負當x＞1時,logax＞0當x＞1時,logax＜0質(zhì)單調(diào)性在（0，＋∞）上是增函數(shù)在（0，＋∞）上是減函數(shù)例1.求下列函數(shù)的定義域：7(1)y=log5x2(2)y=loga(4－x)(3)y=log（5x-1)(7x-2)解∶(1)x2＞0x≠0∴函數(shù)y=logax2的定義域是｛x│x≠0｝(2)4－x＞0x＜4∴函數(shù)y=loga(4－x)的定義域是｛x│x＜4｝(3)要使函數(shù)有意義，必有2

4、x>7x-2>07152522752275例2.比較大?。?①log23log0.71.8③loga4loga3.14當a>1時，loga4>loga3.14當0lo

5、g33=1②因為log32>0log53log53得:log32>log20.8方當?shù)讛?shù)不相同，真數(shù)也不相同時，法練習2：比較大小①log761②log0.531③log671④log0.60.11⑤log35.10⑥log0.120⑦log20.80⑧l(xiāng)og0.20.60例4.比較大?。?1方法③log53log43解:利用對數(shù)函數(shù)圖象得到log531和0

6、為__l_o_g_2_3___>___l_o_g_3_2__>_l__o_g_0_.5_3__.(2)log0.34_<____log0.20.7<>練習5.不等式log2(4x+8)>log22x的解集為（A）A.x>0B.x>-4C.x>-2D.x>4解：由對數(shù)函數(shù)的性質(zhì)及定義域要求，得4x+8>0x>-22x>0X>0∴x>04x+8>2xx>-4解對數(shù)不等式時,注意真數(shù)大于零.解下列不等式：(1)log(x2?3)?log2x(3,3)0.20.2a?1,x?6(2)2log(x?4)?log(x?2)aa0?a?1,4?x?624log[log(x?)]130.25

7、(3)()?122525?1?x??或?x?155求下列函數(shù)的單調(diào)區(qū)間：(1)y?log0.3(x?2)?3(2,??)上遞減(2)y?log(x?2)?5(2,??)上遞增32(?1,1)上遞減(3)y?log(3?2x?x)0.3(1,3)上遞增(3,??)上遞減2(4)y?log(2x?5x?3)10.3(??,?)上遞增2xx?1(5)y?log(4?2?2)0.3(0,??)上遞減,(??,0)上遞增141.回顧對數(shù)函數(shù)研究的過程：課對數(shù)函數(shù)的概念畫出對數(shù)函數(shù)的圖象.觀察圖象特征堂總結(jié)函數(shù)性質(zhì)