3年高考2年模擬2021版新教材高考數(shù)學(xué)第四章指數(shù)函數(shù)對(duì)數(shù)函數(shù)與冪函數(shù)4.2對(duì)數(shù)與對(duì)數(shù)函數(shù)4.2.1對(duì)數(shù)運(yùn)算講義新人教B版必修第二冊(cè)

1、4.2.1對(duì)數(shù)運(yùn)算課標(biāo)解讀課標(biāo)要求核心素養(yǎng)1.理解對(duì)數(shù)的概念,能進(jìn)行指數(shù)式與對(duì)數(shù)式的互化.(重點(diǎn))2.理解對(duì)數(shù)的底數(shù)和真數(shù)的取值范圍.(易混點(diǎn))3.掌握對(duì)數(shù)的基本性質(zhì)及對(duì)數(shù)恒等式.(難點(diǎn))1.通過(guò)對(duì)數(shù)定義及相關(guān)概念的學(xué)習(xí),培養(yǎng)數(shù)學(xué)抽象的核心素養(yǎng).2.通過(guò)對(duì)數(shù)性質(zhì)的學(xué)習(xí),培養(yǎng)數(shù)學(xué)運(yùn)算的核心素養(yǎng).莊子天下篇:一尺之棰,日取其半,萬(wàn)世不竭.意思是一尺長(zhǎng)的木棍,每天截去它的一半,千秋萬(wàn)代也截不完.問(wèn)題1:一尺長(zhǎng)的木棍,取4次,還有多長(zhǎng)?答案124=116(尺).問(wèn)題2:一尺長(zhǎng)的木棍,取多少次,還有0.125尺?請(qǐng)構(gòu)造出方程.答案設(shè)取x次,還有0.125尺,則12x=0.125.1.對(duì)數(shù)的定義及相關(guān)概念

2、(1)對(duì)數(shù)的概念:在表達(dá)式ab=n(a0且a1,n(0,+)中,當(dāng)a與n確定之后,只有唯一的b能滿足這個(gè)式子,此時(shí),冪指數(shù)b稱為以a為底n的對(duì)數(shù),記作b=logan,其中a稱為對(duì)數(shù)的底數(shù),n稱為對(duì)數(shù)的真數(shù).(2)對(duì)數(shù)恒等式:alogan=n.(3)兩類特殊對(duì)數(shù):名稱定義符號(hào)常用對(duì)數(shù)以10為底的對(duì)數(shù)lgn自然對(duì)數(shù)以e為底的對(duì)數(shù)lnn思考:如何理解指數(shù)式與對(duì)數(shù)式的關(guān)系?提示指數(shù)式和對(duì)數(shù)式的關(guān)系如圖所示:2.對(duì)數(shù)的性質(zhì)性質(zhì)1負(fù)數(shù)和零沒(méi)有對(duì)數(shù)性質(zhì)21的對(duì)數(shù)是0,即loga1=0(a0且a1)性質(zhì)3底的對(duì)數(shù)是1,即logaa=1(a0且a1)探究一對(duì)數(shù)的概念例1(易錯(cuò)題)對(duì)數(shù)式log(x-2)(x+2)

3、中實(shí)數(shù)x的取值范圍是.易錯(cuò)辨析:忽視對(duì)數(shù)式中底數(shù)大于0且不等于1,真數(shù)大于0的情況致誤.答案(2,3)(3,+)解析由題意可得x+20,x-20,x-21,解得x2,且x3,所以實(shí)數(shù)x的取值范圍是(2,3)(3,+).易錯(cuò)點(diǎn)撥根據(jù)對(duì)數(shù)的概念,對(duì)數(shù)式的底數(shù)大于0且不等于1,真數(shù)大于0,列出不等式(組),可求得對(duì)數(shù)式中字母的取值范圍.1.對(duì)數(shù)式log(2x-3)(x-1)中實(shí)數(shù)x的取值范圍是.答案32,2(2,+)解析由題意可得x-10,2x-30,2x-31,解得x32,且x2,所以實(shí)數(shù)x的取值范圍是32,2(2,+).探究二指數(shù)式與對(duì)數(shù)式的互化例2(1)將下列指數(shù)式與對(duì)數(shù)式互化:log216=

4、4;43=64;lg1000=3.(2)設(shè)a=log310,b=log37,求3a-b的值.解析(1)因?yàn)閘og216=4,所以24=16.因?yàn)?3=64,所以log464=3.因?yàn)閘g1000=3,所以103=1000.(2)因?yàn)閍=log310,b=log37,所以3a=10,3b=7.所以3a-b=3a3b=107.思維突破指數(shù)式與對(duì)數(shù)式互化的方法技巧(1)指數(shù)式化為對(duì)數(shù)式:將指數(shù)式的冪作為真數(shù),指數(shù)作為對(duì)數(shù),底數(shù)不變,寫出對(duì)數(shù)式.(2)對(duì)數(shù)式化為指數(shù)式:將對(duì)數(shù)式的真數(shù)作為冪,對(duì)數(shù)作為指數(shù),底數(shù)不變,寫出指數(shù)式.2.(1)將下列各等式化為相應(yīng)的對(duì)數(shù)式或指數(shù)式:10-3=11000;ln2

5、=x.(2)已知a0且a1,loga2=m,loga3=n,求a2m+n的值.解析(1)因?yàn)?0-3=11000,所以lg11000=-3.因?yàn)閘n2=x,所以ex=2.(2)由loga2=m得am=2,由loga3=n得an=3,所以a2m+n=a2man=(am)2an=223=12.探究三對(duì)數(shù)的性質(zhì)與對(duì)數(shù)恒等式例3計(jì)算:22+log25=.答案20解析22+log25=222log25=45=20.思維突破解題步驟:3.(1)求2log23+log43的值;(2)已知log2019(log3(log2a)=0,計(jì)算36log6a的值.解析(1)2log23+log43=2log232lo

6、g43=3(4log43)12=33.(2)由log2019(log3(log2a)=0,得log3(log2a)=1,log2a=3,解得a=8,36log6a=(6log68)2=82=64.1.對(duì)數(shù)式x=lg2化為指數(shù)式為()a.10x=2b.x10=2c.x2=10d.2x=10答案a2.若log8x=-23,則x的值為()a.14b.4c.2d.12答案alog8x=-23,x=8-28=2-2=14,故選a.3.2log23=.答案34.若log3(log2x)=0,則x12=.答案2解析log3(log2x)=0,log2x=30=1,x=2,即x12=2.數(shù)據(jù)分析利用恒等轉(zhuǎn)化思

7、想求值(1)設(shè)3a=4b=36,求2a+1b的值;(2)已知2x=3y=5z,且1x+1y+1z=1,求x,y,z的值.素養(yǎng)探究:運(yùn)用數(shù)學(xué)轉(zhuǎn)化思想時(shí)要抓住問(wèn)題的實(shí)質(zhì),在遇到較復(fù)雜的問(wèn)題時(shí),通過(guò)分析問(wèn)題,然后使用一定的方法,使復(fù)雜的問(wèn)題簡(jiǎn)單化,陌生的問(wèn)題熟悉化,抽象的問(wèn)題具體化,過(guò)程中體現(xiàn)數(shù)據(jù)分析核心素養(yǎng).解析(1)由3a=4b=36,得a=log336,b=log436,由換底公式得1a=log363,1b=log364,2a+1b=2log363+log364=log3636=1.(2)令2x=3y=5z=k(k0且k1),則x=log2k,y=log3k,z=log5k,1x=logk2,

8、1y=logk3,1z=logk5,由1x+1y+1z=1,得logk2+logk3+logk5=logk30=1,k=30,x=log230=1+log215,y=log330=1+log310,z=log530=1+log56.已知3a=5b=m,且1a+1b=2,則m=.答案15解析由3a=5b=m,得a=log3m,b=log5m,故1a+1b=logm3+logm5=logm15=2,m=15.課時(shí)達(dá)標(biāo)訓(xùn)練1.下列各式:lg(lg10)=0;lg(lne)=0;若10=lgx,則x=100;若log25x=12,則x=25,其中正確的有()a.1個(gè)b.2個(gè)c.3個(gè)d.4個(gè)答案blg1

9、0=1,lg1=0,所以正確;lne=1,lg1=0,所以正確;10=lgx1010=x,故錯(cuò)誤;log25x=122512=xx=5,故錯(cuò)誤,故選b.2.若logab=c,則()a.ac=bb.ab=cc.ca=bd.cb=a答案a3.使對(duì)數(shù)loga(-2a+1)有意義的a的取值范圍是()a.12,1(1,+)b.0,12c.(0,1)(1,+)d.-,12答案b若使對(duì)數(shù)loga(-2a+1)有意義,則a需滿足a0,a1,-2a+10,解得0a0,b0,若log4a=log6b=12,則ab=.答案63解析由log4a=log6b=12,可得a=412=2,b=612=6,所以ab=26=6

10、3.8.已知log2(log3(log4x)=0,且log4(log2y)=1,求xy34的值.解析log2(log3(log4x)=0,log3(log4x)=1,log4x=3,x=43=64.由log4(log2y)=1,得log2y=4,y=24=16.xy34=641634=88=64.9.(多選)下列指數(shù)式與對(duì)數(shù)式互化正確的是()a.e0=1與ln1=0b.log39=2與912=3c.8-13=12與log812=-13d.log77=1與71=7答案acd對(duì)于a:e0=1可化為0=loge1=ln1,a正確;對(duì)于b:log39=2可化為32=9,b不正確;對(duì)于c:8-13=12

11、可化為log812=-13,c正確;對(duì)于d:log77=1可化為71=7,d正確.故選acd.10.已知logax=2,logbx=1,logcx=4(a,b,c,x0且a,b,c1),則logx(abc)=()a.47b.27c.72d.74答案d易知x=a2=b=c4,所以(abc)4=x7,所以abc=x74,即logx(abc)=74.11.已知函數(shù)f(x)=2x,x4,f(x+2),x4,則f(1+log23)的值為()a.6b.12c.24d.36答案c因?yàn)?322,所以1log232,21+log233,4(1+log23)+20,且a1,loga2=x,則ax=;a2x+a-2

12、x=.答案2;174解析由對(duì)數(shù)式與指數(shù)式的互化,可得loga2=xax=2.a2x+a-2x=(ax)2+1(ax)2=22+122=174.13.已知x2-6x+y2+4y+13=0,則log(x-y)(x+y)的值是.答案0解析由x2-6x+y2+4y+13=0得x2-6x+9+y2+4y+4=0,即(x-3)2+(y+2)2=0,解得x=3,y=-2.故log(x-y)(x+y)=log51=0.14.(1)已知log189=a,log1854=b,求182a-b的值;(2)已知logx27=31+log32,求x的值.解析(1)log189=a,log1854=b,18a=9,18b=54,182a-b=182a