4.2换底公式课后篇巩固提升A组基础巩固1.log49log43的值为()A.12B.2C.32D.92解析:log49log43=log39=2.答案:B2.log9❑√3的值等于()A.4B.14C.-4D.-2解析:log9❑√3=log32312=122log33=14.答案:B3.若10a=2,10b=3,则log36等于()A.a+baB.a+bbC.aa+bD.ba+b解析:由已知得a=lg2,b=lg3,所以log36=lg6lg3=lg2+lg3lg3=a+bb.答案:B4.若a>0,a≠1,x>y>0,n∈N+,则下列各式:①(logax)n=nlogax;②(logax)n=logaxn;③logax=-loga1x;④logaxlogay=logaxy;⑤n√logax=1nlogax;⑥logaxn=logan√x;⑦logaxn=nlogax;⑧logax-yx+y=-logax+yx-y.其中成立的有()A.3个B.4个C.5个D.6个解析:根据对数的运算法则及换底公式得③⑥⑦⑧正确,①②④⑤不正确.答案:B5.已知4a=7,6b=8,则log1221可以用a,b表示为()A.3-b+2ab3+bB.2a+b-ab3+bC.3-b+2ab4-2bD.2a+b-ab4-2b解析:由题意可得a=log47=lg72lg2,则lg7lg2=2a,b=log68=3lg2lg6,则lg6lg2=3b,据此有: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,则n的值为.解析:∵m=1log35=log53,∴n=5m=5log53=3.答案:37.已知3x=4y=36,则2x+1y的值为.解析:∵3x=36,4y=36,∴x=log336,y=log436.∴1x=log363,1y=log364,∴2x+1y=2log363+log364=log36(32×4)=1.答案:18.log35log46log57log68log79=.解析:log35log46log57log68log79=lg5lg3·lg6lg4·lg7lg5·lg8lg6·lg9lg7=lg8lg9lg3lg4=3lg2·2lg3lg3·2lg2=3.答案:39.已知x,y为正数,且3x=4y,求使2x=py成立的p的值.解:设3x=4y=k(显然k≠1),则x=log3k,y=log4k,由2x=py,得2log3k=plog4k=p·log3klog34.∵log3k≠0,∴p=2log34.10.导学号85104068计算:(log43+log83)lg2lg3+log535-2log573+log57-log51.8.解:根据对数的换底公式和运算性质可得(log43+log83)·lg2lg3=(lg3lg4+lg3lg8)·lg2lg3=lg32lg2·lg2lg3+lg33lg2·lg2lg3=12+13=56,log535-2log573+log57-log51.8=log5(5×7)-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组能力提升1.计算log2125·log3132·log519的值为()A.-20B.-5C.5D.20解析:原式=-log225·log332·log59=-lg25lg2·lg32lg3·lg9lg5=-2lg5lg2·5lg2lg3·2lg3lg5=-20.答案:A2.已知f(3x)=1+2x·log23,则f(21007)的值等于()A.2013B.2014C.2015D.2017解析:令3x=t(t>0),则x=log3t,f(t)=1+2·log3t·log23=1+2·lgtlg3·lg3lg2=1+2lgtlg2,所以f(x)=1+2lgxlg2,故f(21007)=1+2lg21007lg2=2015.答案:C3.已知x,y,z都是大于1的正数,m>0,且logxm=24,logym=40,logxyzm=12,则logzm的值为()A.60B.160C.2003D.320解析:由已知logmx=124,logmy=140,logmxyz=112,所以logmx+logmy+logmz=112,即logmz=112−124−140=160,所以logzm=60,故选A.答案:A4.已知2x=3,log483=y,则x+2y=.解析:∵2x=3,∴x=log23.∵log483=y,∴y=log48-log43=log28log24−log23log24=32−12log23,∴x+2y=log23+2(32-12log23)=3.答案:35.导学号85104069(信息题)已知an=log(n+1)(n+2)(n∈N+),观察下列运算:a1a2=log23·log34=2;a1·a2·a3·a4·a5·a6=log23·log34·…·log78=lg3lg2·lg4lg3·lg5lg4·…·lg8lg7=3,……定义使a1·a2·…·ak为整数的k(k∈N+)叫作企盼数.试确定当a1·a2·…·ak=2016时,企盼数k=.解析:a1·a2·…·ak=lg3lg2·lg4lg3·lg5lg4·…·lg(k+2)lg(k+1)=lg(k+2)lg2=log2(k+2)=2016,解得k=22016-2.答案:22016-26.某种汽车安全行驶的稳定性系数μ随使用年数t的变化规律是μ=μ0e-λt,其中μ0,λ是正常数.经检测,当t=2时,μ=0.90μ0,则当稳定性系数降为0.50μ0时,该种汽车已使用的年数为(结果精确到1,参考数据:lg2=0.3010,lg3=0.4771).解析:由0.90μ0=μ0(e-λ)2,得e-λ=❑√0.90.又0.50μ0=μ0(e-λ)t,则12=(❑√0.90)t,两边取常用对数,得lg12=t2lg0.90,故t=2lg21-2lg3=2×0.30101-2×0.4771≈13.答案:137.计算:(1)lg2+lg5-lg8lg50-lg40;(2)lg12-lg58+lg54-log92·log43.解:(1)原式=lg2×58lg5040=lg54lg54=1.(2)(方法一)原式=lg1258+lg54−lg2lg9×lg3lg4=lg(45×54)−lg22lg3×lg32lg2=lg1-14=-14.(方法二)原式=(lg1-lg2)-(lg5-lg8)+(lg5-lg4)-lg2lg9×lg3lg4=-lg2+lg8-lg4-lg22lg3×lg32lg2=-(lg2+lg4)+lg8-14=-lg(2×4)+lg8-14=-14.8.已知log1227=a,求log616的值(用a表示).解:∵由log1227=a,得3lg32lg2+lg3=a,∴lg2=3-a2alg3.∴log616=lg16lg6=4lg2lg2+lg3=4×3-a2a3-a2a+1=12-4a3+a.9.(拓展探究)已知logax+3logxa-logxy=3(a>1).(1)若设x=at,试用a,t表示y;(2)若当01),所以logay=(logax)2-3logax+3,当x=at时,logax=logaat=t,所以logay=t2-3t+3.所以y=at2-3t+3(t≠0).(2)y=a(t-32)2+34,因为01,所以当t=32时,ymin=a34=8,所以a=16,此时x=a32=64.
