4.2.1对数运算4.2.2对数运算法则课后篇巩固提升夯实基础1.若lnx-lny=a,则ln(x2)3-ln(y2)3等于()A.a2B.aC.3a2D.3a答案D解析ln(x2)3-ln(y2)3=3(lnx2-lny2)=3(lnx-ln2-lny+ln2)=3(lnx-lny)=3a.2.已知a>0,a≠1,x>y>0,n∈N+,下列各式:①(logax)n=nlogax;②logax=-loga1x;③logaxlogay=logaxy;④n√logax=1nlogax;⑤1nlogax=logan√x;⑥logax=loganxn;⑦logax-yx+y=-logax+yx-y.其中成立的有()A.3个B.4个C.5个D.6个答案B解析其中②⑤⑥⑦正确.①式中nlogax=logaxn;③式中logaxy=logax-logay;④式中1nlogax=logan√x.3.(多选)已知函数f(x)={log2x,x>0,3x,x≤0.若f(a)=13,则x的可能取值为()A.-1B.❑√2C.3√2D.2答案AC解析当a>0时,由log2a=13,得a=213=3√2,故C正确;当a≤0时,由3a=13,得a=-1,故A正确.4.如果关于lgx的方程lg2x+(lg2+lg3)lgx+lg2lg3=0的两根为lgx1,lgx2,那么x1x2的值为()A.lg2·lg3B.lg2+lg3C.16D.-6答案C解析∵由已知,得lgx1+lgx2=-(lg2+lg3)=-lg6=lg16,又∵lgx1+lgx2=lg(x1x2),∴lg(x1x2)=lg16.∴x1x2=16.5.已知f(x5)=lgx,则f(2)等于()A.lg2B.lg32C.lg132D.15lg2答案D解析(方法一)令x5=2,则x=215,∴f(2)=lg215=15lg2.(方法二)令x5=t,则x=t15,∴原函数可转化为f(t)=lgt15=15lgt,即f(x)=15lgx,∴f(2)=15lg2.6.若2a=3b=6,则1a+1b=()A.2B.3C.12D.1答案D解析∵2a=3b=6,∴a=log26,b=log36.∴1a+1b=1log26+1log36=log62+log63=1.7.若3α=2,则log38-2log36用含a的代数式可表示为()A.a-2B.3a-(1+a)2C.5a-2D.3a-a2答案A解析∵3a=2,∴a=log32,log38-2log36=3log32-2(log33+log32)=log32-2=a-2.8.已知log32=a,则2log36+log30.5=.答案a+2解析原式=2log3(2×3)+log312=2(log32+log33)-log32=log32+2=a+2.9.log56·log67·log78·log89·log910=.答案1lg5解析原式=lg6lg5·lg7lg6·lg8lg7·lg9lg8·lg10lg9=lg10lg5=1lg5.10.若a=log43,则2a+2-a=,1a+1=.答案4❑√33log312解析∵a=log43=log2❑√3,∴2a+2-a=2log2❑√3+2-log2❑√3=❑√3+1❑√3=4❑√33.∵1a=log34,1=log33,∴1a+1=log34+log33=log312.11.已知a,b,c为正数,且lg(ac)lg(bc)+1=0,则lgab的取值范围是.答案(-∞,-2]∪[2,+∞)解析利用对数的运算性质转化为关于lgc的一元二次方程有解问题进行处理.∵由题意,得(lga+lgc)(lgb+lgc)+1=0,∴有(lgc)2+(lga+lgb)lgc+lgalgb+1=0.设lgc=t,则t2+(lga+lgb)t+lgalgb+1=0,t∈R,则关于t的方程t2+(lga+lgb)t+lgalgb+1=0有根,∴Δ=(lga+lgb)2-4(lgalgb+1)≥0.整理,得(lga-lgb)2≥4,∴|lgab|≥2.∴lgab≥2或lgab≤-2,即lgab的取值范围是(-∞,-2]∪[2,+∞).12.计算:log28+lg11000+ln3√e2+21-12log23+(lg5)2+lg2lg50.解原式=3-3+23+2÷212log23+(lg5)2+lg2(lg5+1)=23+2❑√33+(lg5)2+(1-lg5)(1+lg5)=53+2❑√33.能力提升1.设a>0,a≠1,x,y满足logax+3logxa-logxy=3.(1)用logax表示logay;(2)当x取何值时logay取得最小值?解(1)由题意得logax+3logax−logaylogax=3,∴logaylogax=logax+3logax-3.∴logay=(logax)2-3logax+3.(2)设logax=t,t∈R,则有logay=t2-3t+3=(t-32)2+34(t∈R),∴当t=32时,logay取得最小值34,此时logax=32,x=a32,即当x=a32时,logay取得最小值34.2.(1)已知5a=3,5b=4,求a,b,并用a,b表示log2512.(2)求值:21412-(❑√3-π)0+log313+712log74.解(1)因为5a=3,5b=4,所以a=log53,b=log54.所以log2512=log512log525=12(log53+log54)=a+b2.(2)原式=9412-1+(-1)+2=32-1-1+2=32.3.甲、乙两人解关于x的方程log2x+b+clogx2=0,甲写错了常数b,得到两个根14,18;乙写错了常数c得到两个根12,64.求这个方程真正的根.解原方程可化为log2x+b+c·1log2x=0,即(log2x)2+blog2x+c=0.因为甲写错了常数b得到两个根14,18,所以c=log214·log218=6.因为乙写错了常数c得到两个根12,64,所以b=-(log212+log264)=-5.故原方程为(log2x)2-5log2x+6=0.解得log2x=2或log2x=3.所以x=4或x=8,即方程真正的根为4,8.4.已知2y·logy4-2y-1=0,❑√logx❑√5x·log5x=-1,问是否存在一个正整数P,使P=❑√1x-y?解∵2y·logy4-2y-1=0,∴2y(logy4-12)=0.又∵2y>0,∴logy4=12.∴y=16.由❑√logx❑√5x·log5x=-1得❑√logx❑√5x=-logx5>0,∴logx❑√5x=(logx5)2.∴12logx5x=(logx5)2.∴2(logx5)2-logx5-1=0,即(2logx5+1)(logx5-1)=0,∴logx5=-12或logx5=1.∵-logx5>0,∴logx5<0.∴logx5=1(舍去).∴logx5=-12,即x-12=5.∴x=125.∴1x=25.∴P=❑√1x-y=❑√25-16=❑√9=3.即存在正整数P=3,使P=❑√1x-y.
