2022-2023學(xué)年新教材高中數(shù)學(xué)第四章對(duì)數(shù)運(yùn)算與對(duì)數(shù)函數(shù)2對(duì)數(shù)的運(yùn)算2.1對(duì)數(shù)的運(yùn)算性質(zhì)2.2換底公式課后習(xí)題北師大版必修第一冊(cè)

歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！2.1對(duì)數(shù)的運(yùn)算性質(zhì)2.2換底公式A級(jí)必備知識(shí)基礎(chǔ)練1.2log510+log50.25=()A.0 B.1 C.2 D.42.(2022內(nèi)蒙古包頭高三期末(文))若xlog34=1,則3(4x-4-x)=()A.5 B.7 C.8 D.103.1log1A.lg3 B.-lg3 C.1lg3 D.-4.已知2a=3b=k(k≠1),且2a+b=ab,則實(shí)數(shù)k的值為()A.6 B.9 C.12 D.185.(2022江西九江高一期末)設(shè)a=lg2,b=lg3,則log318=()A.2ab+1 B.2ba+1 C.ab+2 6.log35log46log57log68log79=.

7.設(shè)ax=M,y=logaN(a>0,且a≠1,M>0,N>0).試用x,y表示logaM34N58.計(jì)算:(1)lg2+(2)lg12-lg58+lg54-log92·(3)已知log53=a,log54=b,用a,b表示log25144.B級(jí)關(guān)鍵能力提升練9.若lgx-lgy=a,則lgx23-lgy23A.3a B.32a C.a D.10.若2loga(P-2Q)=logaP+logaQ(a>0,且a≠1),則PQ的值為(A.14 B.4 C.1 D.11.(多選題)設(shè)a,b,c都是正數(shù),且4a=6b=9c,那么()A.ab+bc=2ac B.ab+bc=acC.2c=2a12.設(shè)a=log36,b=log520,則log215=()A.a+b-3C.a+2b-313.(2022江西景德鎮(zhèn)一中高一期末(文))已知實(shí)數(shù)x,y,正數(shù)a,b滿(mǎn)足ax=by=2,且2x+1y=-3,則1b14.已知loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1)(a>0,且a≠1),求log8yx的值C級(jí)學(xué)科素養(yǎng)創(chuàng)新練15.設(shè)正數(shù)a,b,c滿(mǎn)足a2+b2=c2.求證:log21+b+ca+log21+a-cb=

2.1對(duì)數(shù)的運(yùn)算性質(zhì)2.2換底公式1.C原式=log5102+log50.25=log5(100×0.25)=log525=2.2.C因?yàn)閤log34=1,所以log34x=1,即4x=3,所以3(4x-4-x)=3×3-13=8.故選C.3.C原式=log1914+log1315=log94+log35=log32+log34.D∵2a=3b=k(k≠1),∴a=log2k,b=log3k,∴1a=logk2,1b=logk∵2a+b=ab,∴2b+1a=2logk3+logk2=logk9+logk2=log∴k=18.5.Clog318=lg18lg3=故選C.6.3log35log46log57log68log79=lg5lg3·lg67.3x-5y4∵ax=M,∴x=loga∴l(xiāng)ogaM34N5=logaM3-loga4N5=3logaM-54log8.解(1)原式=lg2×58lg(2)(方法一)原式=lg1258+lg54?lg2lg9×lg3lg4(方法二)原式=(lg1-lg2)-(lg5-lg8)+(lg5-lg4)-lg2lg9×lg3lg4=-lg2+lg8-lg4-lg22lg3×lg32lg2=-(lg2+lg4)+lg8-14=-lg(2(3)∵log53=a,log54=b,∴l(xiāng)og25144=log512=log53+log54=a+b.9.Algx23-lgy23=3lgx2-lgy210.B由2loga(P-2Q)=logaP+logaQ,得loga(P-2Q)2=loga(PQ),P>0,Q>0,P>2Q.由對(duì)數(shù)運(yùn)算法則得(P-2Q)2=PQ,即P2-5PQ+4Q2=0,所以P=Q(舍去)或P=4Q,解得PQ=411.AD由題意,設(shè)4a=6b=9c=k(k>1),則a=log4k,b=log6k,c=log9k,由ab+bc=2ac,可得bc+因?yàn)閎c+ba=log6klog9k+log6klog2a+1b=2log4k+1log6k=2logk4+logk6=logk96,2b?1a=2log6k?1log4k=2logk6-logk412.D∵a=log36=1+log32,b=log520=1+2log52,∴l(xiāng)og23=1a-1,log25∴l(xiāng)og215=log23+log25=1a-1+13.-132已知實(shí)數(shù)x,y,正數(shù)a,b滿(mǎn)足ax=by=2,則x=loga2,y=logb由換底公式可得2x+1y=2log2a+log2b=log2(a2b)=-3,可得a2b=18,則1因?yàn)閍>0,則1b-a=8a2-a=8a-1162-132≥-1當(dāng)且僅當(dāng)a=116時(shí),等號(hào)成立,因此,1b-a的最小值為-14.解由對(duì)數(shù)的運(yùn)算法則,可將等式化為loga[(x2+4)·(y2+1)]=loga[5(2xy-1)],∴(x2+4)(y2+1)=5(2xy-1).整