課標(biāo)版高考文科數(shù)學(xué)總復(fù)習(xí)專題2.5 對(duì)數(shù)與對(duì)數(shù)函數(shù)（講解練）教學(xué)講練

1、專題二函數(shù)2.5對(duì)數(shù)與對(duì)數(shù)函數(shù)高考文數(shù)考點(diǎn)一對(duì)數(shù)的概念及運(yùn)算考點(diǎn)清單考向基礎(chǔ)1.對(duì)數(shù)的概念(1)對(duì)數(shù)的定義一般地,如果ax=N(a0且a1),那么數(shù)x叫做以a為底N的對(duì)數(shù),記作x=logaN,其中a叫做對(duì)數(shù)的底數(shù),N叫做真數(shù).(2)幾種常見(jiàn)對(duì)數(shù)對(duì)數(shù)形式特點(diǎn)記法一般對(duì)數(shù)底數(shù)為a(a0且a1)logaN常用對(duì)數(shù)底數(shù)為10lg N自然對(duì)數(shù)底數(shù)為eln N性質(zhì)loga1=0;logaa=1=N;logaaN=N(a0且a1)換底公式logbN=(a,b均大于0且不等于1,N0)相關(guān)結(jié)論:logab=;logablogbclogcd=logad(a,b,c均大于0且不等于1,d0)運(yùn)算法則條件a0且a1

2、,M0,N0結(jié)論loga(MN)=logaM+logaNloga=logaM-logaNlogaMn=nlogaM(nR)2.對(duì)數(shù)的性質(zhì)、換底公式與運(yùn)算法則考向?qū)?shù)的運(yùn)算考向突破例1(2018皖西高中教學(xué)聯(lián)盟期末,4)計(jì)算log29log34+2log510+log50.25=()A.0B.2C.4D.6解析由對(duì)數(shù)的運(yùn)算公式和換底公式可得log29log34+2log510+log50.25=2log23+log5(1020.25)=4+2=6.故選D.答案D考向基礎(chǔ)1.對(duì)數(shù)函數(shù)的圖象與性質(zhì) a10a1時(shí),y0;當(dāng)0 x1時(shí),y1時(shí),y0;當(dāng)0 x0是(0,+)上的增函數(shù)是(0,+)上的減函數(shù)

3、考點(diǎn)二對(duì)數(shù)函數(shù)的圖象與性質(zhì)2.反函數(shù)指數(shù)函數(shù)y=ax(a0,且a1)與對(duì)數(shù)函數(shù)y=logax(a0且a1)互為反函數(shù),它們的圖象關(guān)于直線y=x對(duì)稱.其圖象關(guān)系如圖所示.由圖象可知,ab1cd,在第一象限內(nèi),從左向右,底數(shù)越來(lái)越大.3.比較底數(shù)的大小考向突破考向一對(duì)數(shù)函數(shù)圖象的應(yīng)用例2(2020屆山西運(yùn)城模擬,7)已知函數(shù)f(x)=|ln x|滿足f(a)f(2-a),則實(shí)數(shù)a的取值范圍是()A.(0,1)B.(1,2)C.(2,3)D.(1,3)解析f(x)=畫出f(x)的大致圖象如圖,由圖知f(x)在(0,1)上單調(diào)遞減,在(1,+)上單調(diào)遞增.根據(jù)題意可知0a2.當(dāng)0a1時(shí),f(a)f(2

4、-a),-ln aln(2-a)a(2-a)1,解得a10a1;當(dāng)a=1時(shí),f(a)=f(2-a),不符合題意;當(dāng)1a2,02-af(2-a),ln a-ln(2-a)a(2-a)1,無(wú)解.綜上,a的取值范圍為(0,1),故選A.答案A例3(2019陜西西安高新區(qū)第一中學(xué)模擬,6)已知函數(shù)f(x)=5-log3x,x(3,27,則f(x)的值域是()A.(2,4 B.2,4)C.-4,4)D.(6,9考向二對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用解析因?yàn)?x27,所以1log3x3,2f(x)0的x的值組成的集合.(2)先確定f(x)0時(shí)對(duì)應(yīng)的x的取值范圍及此時(shí)f(x)的取值范圍,再根據(jù)對(duì)數(shù)函數(shù)的單調(diào)性確定y=lo

5、ga f(x)的值域.2.與對(duì)數(shù)函數(shù)有關(guān)的復(fù)合函數(shù)的單調(diào)性函數(shù)y=loga f(x)的單調(diào)區(qū)間必須保證在f(x)0時(shí)相應(yīng)x的取值范圍內(nèi),這時(shí)內(nèi)外層函數(shù)要注意“同增異減”.考向一與對(duì)數(shù)函數(shù)有關(guān)復(fù)合函數(shù)的值域考向突破例4(2018江西一模,15)若函數(shù)f(x)=loga(a0且a1)的值域?yàn)镽,則實(shí)數(shù)a的取值范圍是.解析函數(shù)f(x)=loga(a0且a1)的值域?yàn)镽,則x+-4能取到所有正數(shù).易知x0,x+2,只需2-40,即24,解得a4.故實(shí)數(shù)a的取值范圍是(0,1)(1,4.答案(0,1)(1,4考向二與對(duì)數(shù)函數(shù)有關(guān)的復(fù)合函數(shù)的單調(diào)性例5函數(shù)f(x)=log2(x2-2x-3)的單調(diào)增區(qū)間是

6、.解析由題意可知x2-2x-30,x3或x-1.令u=x2-2x-3,該函數(shù)在(-,-1)上單調(diào)遞減,在(3,+)上單調(diào)遞增,又y=log2u在(0,+)上單調(diào)遞增,y=log2(x2-2x-3)在(-,-1)上單調(diào)遞減,在(3,+)上單調(diào)遞增,故f(x)的單調(diào)增區(qū)間為(3,+).答案(3,+)例6(2019山西呂梁第一次模擬,6)已知a=log35,b=1.51.5,c=ln 2,則a,b,c的大小關(guān)系是()A.cabB.cbaC.acbD.abc考向三指數(shù)式與對(duì)數(shù)式的大小比較解析1a=log35=log325 1.5,c=ln 21,所以ca1時(shí),圖象上升;0a1,b1(或0a1,0b1時(shí)

7、,“底大圖低”,即若 ab,則y1y2;當(dāng)0 xb,則y1y2.3.對(duì)一些可通過(guò)平移、對(duì)稱作出其圖象的對(duì)數(shù)函數(shù)型問(wèn)題,在求解其單調(diào)性(單調(diào)區(qū)間)、值域(最值)、零點(diǎn)時(shí),常利用數(shù)形結(jié)合法.方法技巧例1(2018廣東廣州執(zhí)信中學(xué)月考,5)設(shè)a,c為正數(shù),且3a=loa,=9,=log3c,則()A.bacB.cbaC.cabD.abc解析方程的根可以轉(zhuǎn)化為兩圖象交點(diǎn)的橫坐標(biāo),a為y=3x與y=lox兩函數(shù)圖象交點(diǎn)的橫坐標(biāo),c為y=與y=log3x兩函數(shù)圖象交點(diǎn)的橫坐標(biāo),易得b=-2.畫出y=,y=3x,y=log3x,y=lox的圖象,可看出ba0且a1).例2(2020屆河北邯鄲模擬,15)已知函數(shù)f(x)=|log3x|,實(shí)數(shù)m,n滿足0mn,且f(m)=f(n),若f(x)在m2,n上的最大值為2,則=.解析f(x)=|log3x|,正實(shí)數(shù)m,n滿足mn,且f(m)=f(n),0m1n,-log3m=log3n,mn=1.f(x)在區(qū)間m2,n上的最大值為2,函數(shù)f(x)在m2,1上是減函數(shù),