復(fù)習(xí)對(duì)數(shù)函數(shù)及簡(jiǎn)單對(duì)數(shù)方程

對(duì)數(shù)函數(shù)及其性質(zhì)復(fù)習(xí)對(duì)數(shù)函數(shù)及簡(jiǎn)單對(duì)數(shù)方程一．復(fù)習(xí)對(duì)數(shù)函數(shù)1．對(duì)數(shù)函數(shù)的定義2．對(duì)數(shù)函數(shù)的圖象與性質(zhì),通過(guò)圖象確定底數(shù)大小

練習(xí)：1．比較大小2．對(duì)數(shù)不等式二：簡(jiǎn)單對(duì)數(shù)方程1．對(duì)數(shù)方程的定義2．解對(duì)數(shù)方程三：小結(jié)四：作業(yè)xyo1定義域X(0,+)值域R

R單調(diào)性奇偶性過(guò)定點(diǎn)0<x<1X>1單調(diào)遞減單調(diào)遞增非奇非偶非奇非偶(1,0)(1,0)

y>0y<0y<0y>0圖象0<a<1a>1

1xy0x(0,+)1oxyxyo1a1a3a2a1a2a3y=logax0<a<1a>1比較底數(shù)a1<a2<a3a1<a2<a3

圖象1oyx

a1a2a3a1a2a31yxo

例１比較大?。孩賚og23log23.5②

log0.71.6log0.71.8③

loga4loga3.14④

log35log54<>

③loga4loga3.14解:討論a的情況I.當(dāng)0<a<1時(shí)

y=logax是減函數(shù)因?yàn)?>3.14所以loga4<loga3.14II.當(dāng)a>1時(shí)y=logax是增函數(shù)因?yàn)?>3.14所以loga4>loga3.14

解:因?yàn)閘og35>1,log54<1得:log35>log54④log35log54例1(5)log56log47解:

利用對(duì)數(shù)函數(shù)圖象y1=log4xy2=log5x7xoy由函數(shù)單調(diào)性log56<log57

插入中間量log57（或log46)再比較log57與log47的大小所以log56>log47得到log57<log47例2若loga0.75>1求a取值范圍解：loga0.75>logaa根據(jù)y=logax的單調(diào)性進(jìn)行討論I0<a<10.75<a得0.75<a<1IIa>10.75>a得a由I、II得0.75<a<1例3解不等式

(1)log2(x2-4x+8)>log22x

X2-4x+8>0解2x>0X2-4x+8>2xxRX>0X<2或x>40<x<2或

x>4

(2)log2(log0.5x)>1解:log2(log0.5x)>log22log0.5x>0log0.5x>2X>0X<1X<0.250<x<0.25例3解不等式log2(x+4)+log2(x-1)=1+log2(x+8)解:log2(x+4)(x-1)=log22(x+8)(x+4)(x+1)=2(x+8)X2+x-20=0x=-5或

x=4經(jīng)檢驗(yàn)x=-5(舍去)原方程的解為x=4例4解方程

例5解方程xlgx=1000x2解:lg(xlgx)=lg(1000x2)lgxlgx=3+2lgx(lgx)2-2lgx-3=0令lgx=tt2-2t-3=0t1=３或

t2=-1lgx=3或

lgx=-1x1=1000x2=0.1經(jīng)檢驗(yàn)x1=1000,x2=0.1是原方程的解

例6解方程logx3+logx+13=0解:

小結(jié)1．應(yīng)用對(duì)數(shù)函數(shù)的圖象與性質(zhì)，