/04/4/對(duì)數(shù)運(yùn)算法則必備知識(shí)基礎(chǔ)練1.已知a>0,a≠1,x>y>0,n∈N+,下列各式:①(logax)n=nlogax;②logax=-loga;③=loga;④logax;⑤logax=loga;⑥logax=loxn;⑦loga=-loga.其中成立的有()A.3個(gè) B.4個(gè) C.5個(gè) D.6個(gè)答案B解析其中②⑤⑥⑦正確.①式中nlogax=logaxn;③式中l(wèi)oga=logax-logay;④式中l(wèi)ogax=loga.2.等于()A.lg3B.-lgB.-lg3C. D.-答案C解析原式=lo+lo=log94+log35=log32+log35=log310=.3.(多選題)(2020山東臨沂高三期末)若10a=4,10b=25,則下列結(jié)論正確的是()A.a+b=2 B.b-a=1C.ab>8lg22 D.b-a>lg6答案ACD解析由10a=4,10b=25,得a=lg4,b=lg25,∴a+b=lg4+lg25=lg100=2,∴b-a=lg25-lg4=lg.∵lg>lg6,∴b-a>lg6.∴ab=4lg2lg5>4lg2lg4=8lg22.4.計(jì)算:2+lg4+2lg5-eln3=.?答案2解析原式=(33+(lg4+lg25)-eln3=3+2-3=2.5.若a=log43,則2a+2-a=,+1=.?答案log312解析∵a=log43=log2,∴2a+2-a=.∵=log34,1=log33,∴+1=log34+log33=log312.6.計(jì)算:(1);(2)log28+lg+ln+(lg5)2+lg2lg50.解(1)原式==1.(2)原式=3-3++2÷+(lg5)2+lg2(lg5+1)=+lg5(lg5+lg2)+lg2=.關(guān)鍵能力提升練7.若2loga(P-2Q)=logaP+logaQ(a>0且a≠1),則的值為()A. B.4C.1 D.4或1答案B解析由2loga(P-2Q)=logaP+logaQ,得loga(P-2Q)2=loga(PQ).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=4.8.(2021河南高二期末)已知log47=a,4b=6,則log4228=()A. B. C. D.答案D解析由4b=6,得log46=b,因?yàn)閘og47=a,所以log4228=.故選D.9.(多選題)設(shè)a,b,c都是正數(shù),且4a=6b=9c,則下列結(jié)論正確的是()A.ab+bc=2ac B.ab+bc=acC. D.答案AD解析由題意,設(shè)4a=6b=9c=k(k>0),則a=log4k,b=log6k,c=log9k,對(duì)于選項(xiàng)A,由ab+bc=2ac,可得=2,因?yàn)?log69+log64=log636=2,故A正確,B錯(cuò)誤;對(duì)于選項(xiàng)C,=2logk4+logk6=logk96,=2logk9=logk81,故,即C錯(cuò)誤;對(duì)于選項(xiàng)D,=2logk6-logk4=logk9,=logk9,故,即D正確.10.2x=5y=m(m>0),且=2,則m的值為.?答案解析由2x=5y=m(m>0),得x=log2m,y=log5m,由=2,得=2,即logm2+logm5=2,logm(2×5)=2.故有m=.11.方程log2(9x-1-5)=log2(3x-1-2)+2的解為.?答案x=2解析∵log2(9x-1-5)=log2(3x-1-2)+2,∴l(xiāng)og2(9x-1-5)=log2[4×(3x-1-2)],∴9x-1-5=4(3x-1-2),化為(3x)2-12·3x+27=0,因式分解為(3x-3)(3x-9)=0,∴3x=3或3x=9,解得x=1或x=2.經(jīng)過(guò)驗(yàn)證x=1不滿足條件,舍去.∴x=2.12.甲、乙兩人解關(guān)于x的方程log2x+b+clogx2=0,甲寫錯(cuò)了常數(shù)b,得到兩個(gè)根;乙寫錯(cuò)了常數(shù)c得到兩個(gè)根,64.求這個(gè)方程真正的根.解原方程可化為log2x+b+c·=0,即(log2x)2+blog2x+c=0.因?yàn)榧讓戝e(cuò)了常數(shù)b得到兩個(gè)根,所以c=log2×log2=6.因?yàn)橐覍戝e(cuò)了常數(shù)c得到兩個(gè)根,64,所以b=-=-5.故原方程為(log2x)2-5log2x+6=0.解得log2x=2或log2x=3.所以x=4或x=8,即方程真正的根為4,8.學(xué)科素養(yǎng)創(chuàng)新練13.已知2y·logy4-2y-1=0,·log5x=-1,是否存在一個(gè)正整數(shù)P,使P=?解存在.理由如下,∵2y·logy4-2y-1=0,∴2y=0.又2y>0,∴l(xiāng)ogy4=.∴y=16.由·log5x=-1得=-logx5>0,∴l(xiāng)ogx=(logx5)2.∴l(xiāng)ogx5x=(logx5)2.∴2(logx5)2-logx5-1=0,即(2log