2022-2023学年新教材高中数学 第四章 对数运算与对数函数 2 对数的运算 2.docx

1、2.1对数的运算性质2.2换底公式A级必备知识基础练1.2log510+log50.25=()A.0B.1C.2D.42.(2022内蒙古包头高三期末(文)若xlog34=1,则3(4x-4-x)=()A.5B.7C.8D.103.1log1419+1log1513等于()A.lg 3B.-lg 3C.1lg3D.-1lg34.已知2a=3b=k(k1),且2a+b=ab,则实数k的值为()A.6B.9C.12D.185.(2022江西九江高一期末)设a=lg 2,b=lg 3,则log318=()A.2ab+1B.2ba+1C.ab+2D.ba+26.log35log46log57log68

2、log79=.7.设ax=M,y=logaN(a0,且a1,M0,N0).试用x,y表示logaM34N5=.8.计算:(1)lg2+lg5-lg8lg50-lg40;(2)lg12-lg58+lg54-log92log43;(3)已知log53=a,log54=b,用a,b表示log25144.B级关键能力提升练9.若lg x-lg y=a,则lgx23-lgy23=()A.3aB.32aC.aD.a210.若2loga(P-2Q)=logaP+logaQ(a0,且a1),则PQ的值为()A.14B.4C.1D.4或111.(多选题)设a,b,c都是正数,且4a=6b=9c,那么()A.ab

3、+bc=2acB.ab+bc=acC.2c=2a+1bD.1c=2b1a12.设a=log36,b=log520,则log215=()A.a+b-3(a-1)(b-1)B.a+b-2(a-1)(b-1)C.a+2b-3(a-1)(b-1)D.2a+b-3(a-1)(b-1)13.(2022江西景德镇一中高一期末(文)已知实数x,y,正数a,b满足ax=by=2,且2x+1y=-3,则1b-a的最小值为.14.已知loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1)(a0,且a1),求log8yx的值.C级学科素养创新练15.设正数a,b,c满足a2+b2=c2.求证:

4、log21+b+ca+log21+a-cb=1.2.1对数的运算性质2.2换底公式1.C原式=log5102+log50.25=log5(1000.25)=log525=2.2.C因为xlog34=1,所以log34x=1,即4x=3,所以3(4x-4-x)=33-13=8.故选C.3.C原式=log1914+log1315=log94+log35=log32+log35=log310=1lg3.4.D2a=3b=k(k1),a=log2k,b=log3k,1a=logk2,1b=logk3.2a+b=ab,2b+1a=2logk3+logk2=logk9+logk2=logk18=1,k=1

5、8.5.Clog318=lg18lg3=lg2+lg32lg3=a+2bb=ab+2,故选C.6.3log35log46log57log68log79=lg5lg3lg6lg4lg7lg5lg8lg6lg9lg7=lg8lg9lg3lg4=3lg22lg3lg32lg2=3.7.3x-5y4ax=M,x=logaM,logaM34N5=logaM3-loga4N5=3logaM-54logaN=3x-54y.8.解(1)原式=lg258lg5040=lg54lg54=1.(2)(方法一)原式=lg1258+lg54lg2lg9lg3lg4=lg4554lg22lg3lg32lg2=lg1-14

6、=-14.(方法二)原式=(lg1-lg2)-(lg5-lg8)+(lg5-lg4)-lg2lg9lg3lg4=-lg2+lg8-lg4-lg22lg3lg32lg2=-(lg2+lg4)+lg8-14=-lg(24)+lg8-14=-14.(3)log53=a,log54=b,log25144=log512=log53+log54=a+b.9.Algx23-lgy23=3lgx2-lgy2=3(lgx-lgy)=3a.10.B由2loga(P-2Q)=logaP+logaQ,得loga(P-2Q)2=loga(PQ),P0,Q0,P2Q.由对数运算法则得(P-2Q)2=PQ,即P2-5PQ+

7、4Q2=0,所以P=Q(舍去)或P=4Q,解得PQ=4.11.AD由题意,设4a=6b=9c=k(k1),则a=log4k,b=log6k,c=log9k,由ab+bc=2ac,可得bc+ba=2,因为bc+ba=log6klog9k+log6klog4k=logk9logk6+logk4logk6=log69+log64=log636=2,故A正确,B错误;2a+1b=2log4k+1log6k=2logk4+logk6=logk96,2c=2log9k=2logk9=logk81,故2c2a+1b,故C错误;2b1a=2log6k1log4k=2logk6-logk4=logk9,1c=1

8、log9k=logk9,故1c=2b1a,故D正确.12.Da=log36=1+log32,b=log520=1+2log52,log23=1a-1,log25=2b-1,log215=log23+log25=1a-1+2b-1=2a+b-3(a-1)(b-1).故选D.13.-132已知实数x,y,正数a,b满足ax=by=2,则x=loga2,y=logb2,由换底公式可得2x+1y=2log2a+log2b=log2(a2b)=-3,可得a2b=18,则1b=8a2,因为a0,则1b-a=8a2-a=8a-1162-132-132,当且仅当a=116时,等号成立,因此,1b-a的最小值为-132.14.解由对数的运算法则,可将等式化为loga(x2+4)(y2+1)=loga5(2xy-1),(x2+4)(y2+1)=5(2xy-1).整理,得x2y2+x2+4y2-10xy+9=0,配方,得(xy-3)2+(x-2y)2=0,xy=3,x=2y.yx=12.log8yx=log812=log232-1=-13log22=-13.15.证明log21+b+ca+log21+a-cb=log21+b+ca1+a-cb=log2(a+b+c)(a+b-c)ab=log2(a+b)2-c2ab=log22=1.