高中數(shù)學(xué)必修1配北師版-課后習(xí)題及答案3.4.2

1、4.2換底公式課后篇鞏固提升A組基礎(chǔ)鞏固1.log49log43的值為()A.12B.2C.32D.92解析:log49log43=log39=2.答案:B2.log93的值等于()A.4B.14C.-4D.-2解析:log93=log32312=122log33=14.答案:B3.若10a=2,10b=3,則log36等于()A.a+baB.a+bbC.aa+bD.ba+b解析:由已知得a=lg 2,b=lg 3,所以log36=lg6lg3=lg2+lg3lg3=a+bb.答案:B4.若a0,a1,xy0,nN+,則下列各式:(logax)n=nlogax;(logax)n=logaxn;

2、logax=-loga1x;logaxlogay=logaxy;nlogax=1nlogax;logaxn=loganx;logaxn=nlogax;logax-yx+y=-logax+yx-y.其中成立的有()A.3個B.4個C.5個D.6個解析:根據(jù)對數(shù)的運算法則及換底公式得正確,不正確.答案:B5.已知4a=7,6b=8,則log1221可以用a,b表示為()A.3-b+2ab3+bB.2a+b-ab3+bC.3-b+2ab4-2bD.2a+b-ab4-2b解析:由題意可得a=log47=lg72lg2,則lg7lg2=2a,b=log68=3lg2lg6,則lg6lg2=3b,據(jù)此有:

3、log1221=lg21lg12=lg6+lg7-lg2lg6+lg2=lg6lg2+lg7lg2-1lg6lg2+1=3b+2a-13b+1=3+2ab-b3+b.答案:A6.若mlog35=1,n=5m,則n的值為.解析:m=1log35=log53,n=5m=5log53=3.答案:37.已知3x=4y=36,則2x+1y的值為.解析:3x=36,4y=36,x=log336,y=log436.1x=log363,1y=log364,2x+1y=2log363+log364=log36(324)=1.答案:18.log35log46log57log68log79=.解析:log35log

4、46log57log68log79=lg5lg3lg6lg4lg7lg5lg8lg6lg9lg7=lg8lg9lg3lg4=3lg22lg3lg32lg2=3.答案:39.已知x,y為正數(shù),且3x=4y,求使2x=py成立的p的值.解:設(shè)3x=4y=k(顯然k1),則x=log3k,y=log4k,由2x=py,得2log3k=plog4k=plog3klog34.log3k0,p=2log34.10.導(dǎo)學(xué)號85104068計算:(log43+log83)lg2lg3+log535-2log573+log57-log51.8.解:根據(jù)對數(shù)的換底公式和運算性質(zhì)可得(log43+log83)lg2

5、lg3=lg3lg4+lg3lg8lg2lg3=lg32lg2lg2lg3+lg33lg2lg2lg3=12+13=56,log535-2log573+log57-log51.8=log5(57)-2(log57-log53)+log57-log595=1+log57-2log57+2log53+log57-2log53+1=2,所以(log43+log83)lg2lg3+log535-2log573+log57-log51.8=176.B組能力提升1.計算log2125log3132log519的值為()A.-20B.-5C.5D.20解析:原式=-log225log332log59=-lg

6、25lg2lg32lg3lg9lg5=-2lg5lg25lg2lg32lg3lg5=-20.答案:A2.已知f(3x)=1+2xlog23,則f(21 007)的值等于()A.2 013B.2 014C.2 015D.2 017解析:令3x=t(t0),則x=log3t,f(t)=1+2log3tlog23=1+2lgtlg3lg3lg2=1+2lgtlg2,所以f(x)=1+2lgxlg2,故f(21 007)=1+2lg 21 007lg2=2 015.答案:C3.已知x,y,z都是大于1的正數(shù),m0,且logxm=24,logym=40,logxyzm=12,則logzm的值為()A.6

7、0B.160C.2003D.320解析:由已知logmx=124,logmy=140,logmxyz=112,所以logmx+logmy+logmz=112,即logmz=112124140=160,所以logzm=60,故選A.答案:A4.已知2x=3,log483=y,則x+2y=.解析:2x=3,x=log23.log483=y,y=log48-log43=log28log24log23log24=3212log23,x+2y=log23+232-12log23=3.答案:35.導(dǎo)學(xué)號85104069(信息題)已知an=log(n+1)(n+2)(nN+),觀察下列運算:a1a2=log

8、23log34=2;a1a2a3a4a5a6=log23log34log78=lg3lg2lg4lg3lg5lg4lg8lg7=3,定義使a1a2ak為整數(shù)的k(kN+)叫作企盼數(shù).試確定當(dāng)a1a2ak=2 016時,企盼數(shù)k=.解析:a1a2ak=lg3lg2lg4lg3lg5lg4lg(k+2)lg(k+1)=lg(k+2)lg2=log2(k+2)=2 016,解得k=22 016-2.答案:22 016-26.某種汽車安全行駛的穩(wěn)定性系數(shù)隨使用年數(shù)t的變化規(guī)律是=0e-t,其中0,是正常數(shù).經(jīng)檢測,當(dāng)t=2時,=0.900,則當(dāng)穩(wěn)定性系數(shù)降為0.500時,該種汽車已使用的年數(shù)為(結(jié)果精

9、確到1,參考數(shù)據(jù):lg 2=0.301 0,lg 3=0.477 1).解析:由0.900=0(e-)2,得e-=0.90.又0.500=0(e-)t,則12=(0.90)t,兩邊取常用對數(shù),得lg12=t2lg 0.90,故t=2lg21-2lg3=20.301 01-20.477 113.答案:137.計算:(1)lg2+lg5-lg8lg50-lg40;(2)lg12-lg58+lg54-log92log43.解:(1)原式=lg258lg5040=lg54lg54=1.(2)(方法一)原式=lg1258+lg54lg2lg9lg3lg4=lg4554lg22lg3lg32lg2=lg

10、1-14=-14.(方法二)原式=(lg 1-lg 2)-(lg 5-lg 8)+(lg 5-lg 4)-lg2lg9lg3lg4=-lg 2+lg 8-lg 4-lg22lg3lg32lg2=-(lg 2+lg 4)+lg 8-14=-lg (24)+lg 8-14=-14.8.已知log1227=a,求log616的值(用a表示).解:由log1227=a,得3lg32lg2+lg3=a,lg 2=3-a2alg 3.log616=lg16lg6=4lg2lg2+lg3=43-a2a3-a2a+1=12-4a3+a.9.(拓展探究)已知logax+3logxa-logxy=3(a1).(1)若設(shè)x=at