對(duì)數(shù)的運(yùn)算性質(zhì)導(dǎo)學(xué)案 高一上學(xué)期數(shù)學(xué)北師大版（2019）必修+第一冊(cè)

§2對(duì)數(shù)的運(yùn)算2.1對(duì)數(shù)的運(yùn)算性質(zhì)【學(xué)習(xí)目標(biāo)】1.掌握對(duì)數(shù)的運(yùn)算性質(zhì).2.理解對(duì)數(shù)運(yùn)算性質(zhì)的推導(dǎo)過程.3.通過推導(dǎo)對(duì)數(shù)運(yùn)算性質(zhì)的過程,提升數(shù)學(xué)運(yùn)算的核心素養(yǎng).◆知識(shí)點(diǎn)對(duì)數(shù)的運(yùn)算性質(zhì)若a>0,且a≠1,M>0,N>0,b∈R,則有(1)loga(M·N)=;

(2)logaMN=(3)logaMb=blogaM.【診斷分析】判斷正誤.(請(qǐng)?jiān)诶ㄌ?hào)中打“√”或“×”)(1)loga(M·N)=logaM·logaN,其中a>0,且a≠1,M>0,N>0. ()(2)loga(M±N)=logaM±logaN,其中a>0,且a≠1,M>0,N>0. ()(3)logaM·logaN=loga(M+N),其中a>0,且a≠1,M>0,N>0. ()(4)log214=-2. (◆探究點(diǎn)一對(duì)數(shù)的運(yùn)算性質(zhì)例1已知a>0且a≠1,x>0,y>0,z>0,用logax,logay,logaz表示下列各式:(1)loga(xyz);(2)logaxy(3)loga3x2yz;變式[2024·廣東兩陽(yáng)中學(xué)高一月考]若2a=log28,b=log213,則1a+1b= ()A.0 B.2 C.4 D.8[素養(yǎng)小結(jié)]利用對(duì)數(shù)的運(yùn)算性質(zhì)解題時(shí),一定要注意真數(shù)的取值范圍是(0,+∞).公式可以逆用.◆探究點(diǎn)二利用對(duì)數(shù)運(yùn)算性質(zhì)求值例2計(jì)算:(1)4lg2+3lg5-lg15(2)lg27(3)2log32-log3329+log38-5變式計(jì)算:(1)12lg3249-43lg8+(2)lg2+lg5-lg8lg50(3)lg5·(lg8+lg1000)+(lg23)2+lg16+lg0.[素養(yǎng)小結(jié)]利用對(duì)數(shù)的運(yùn)算性質(zhì)求值時(shí),一般有兩個(gè)思路:(1)正用公式:將式中真數(shù)的積、商、冪運(yùn)用對(duì)數(shù)的運(yùn)算性質(zhì)化為對(duì)數(shù)的和、差、積,然后化簡(jiǎn)求值;(2)逆用公式:將式中對(duì)數(shù)的和、差、積運(yùn)用對(duì)數(shù)的運(yùn)算性質(zhì)化為真數(shù)的積、商、冪,然后化簡(jiǎn)求值.拓展已知m>0,且10x=lg(10m)+lg1m,求實(shí)數(shù)x的值◆探究點(diǎn)三條件求值與化簡(jiǎn)例3(1)已知lg2=m,lg3=n,求100m-2n的值.(2)已知log32=a,3b=5,用a,b表示log330.變式(1)設(shè)5m=9,則log53= ()A.2m B.1m C.m D(2)已知log53=a,log54=b,則log5270可表示為 ()A.32ab B.3a+b2C.3a+b2 D.a3+b+[素養(yǎng)小結(jié)]條件求值與化簡(jiǎn)的基本原則和方法:①正用或逆用對(duì)數(shù)運(yùn)算性質(zhì);②對(duì)真數(shù)進(jìn)行處理,要么把已知條件的真數(shù)化簡(jiǎn),要么化簡(jiǎn)所求對(duì)數(shù)的真數(shù).§2對(duì)數(shù)的運(yùn)算2.1對(duì)數(shù)的運(yùn)算性質(zhì)【課前預(yù)習(xí)】知識(shí)點(diǎn)(1)logaM+logaN(2)logaM-logaN診斷分析(1)×(2)×(3)×(4)√【課中探究】探究點(diǎn)一例1解:(1)loga(xyz)=logax+logay+logaz.(2)logaxy2z=loga(xy2)-logaz=logax+logay2-logaz=logax+2logay-(3)loga3x2yz=loga(3x2y)-logaz=logax23+logay-logaz12=2(4)logax5y3z2=loga(x5y3)-logaz2=logax5+logay3-2logaz=5logax+3log變式A[解析]由2a=log28得2a=3,所以a=log23,又b=log213,所以a+b=log23+log213=log21=0,所以1a+1b=a+b探究點(diǎn)二例2解:(1)原式=lg24×5315=lg(2)原式=lg(33)1(3)原式=2log32-(log332-log39)+3log32-3=5log32-(5log32-2)-3=-1.變式解:(1)原式=12(5lg2-2lg7)-43×32lg2+12(2lg7+lg5)=52lg2-lg7-2lg2+lg7+12lg5=12lg2+12lg5=12(lg2+lg(2)原式=lg2×58lg5040+log2(2)-1=(3)原式=3lg5·(1+lg2)+3(lg2)2-lg6+lg6-2=3lg2·(lg5+lg2)+3lg5-2=3-2=1.拓展解:10x=lg(10m)+lg1m=lg10m·1m=lg10=探究點(diǎn)三例3解:(1)∵lg2=m,lg3=n,∴10m=2,10n=3,∴100m-2n=(102)m-2n=102m-4n=102m104n(2)∵3b=5,∴b=log35.又log32=a,∴l(xiāng)og330=12log3(2×3×5)=12(log32+log33+log35)=12(變式(