.3對(duì)數(shù)4.3.1對(duì)數(shù)的概念4.3.2對(duì)數(shù)的運(yùn)算基礎(chǔ)過關(guān)練題組一對(duì)數(shù)的概念及性質(zhì)1.將23=8化為對(duì)數(shù)式正確的是()A.log23=8B.log28=3C.log82=3D.log32=82.(多選題)下列指數(shù)式與對(duì)數(shù)式互化正確的是()A.e3=m與log

3m=eB.10x=6與lgC.27-13=13與log1327=-13D.3.已知log7[log2(lgx)]=0,則x=.

4.已知log32=a,則9a+9-a的值為.

題組二對(duì)數(shù)的運(yùn)算性質(zhì)及對(duì)數(shù)式的恒等變形5.以下四個(gè)結(jié)論中正確的是()A.log28=4B.log35+log34=2C.lg(lg10)=0D.4lo6.(教材習(xí)題改編)2log62+3log633A.log6237.計(jì)算:eln2+0.125-23+log

8.計(jì)算:log512-1+9.計(jì)算:lg52+23lg8+lg5×lg20+(lg2)2=10.(易錯(cuò)題)若f(x)=4x,則f(log23)=.

11.(1)已知lg2=m,lg3=n,試用m,n表示log512;(2)計(jì)算:3log32-23log23×log38+13題組三對(duì)數(shù)運(yùn)算的應(yīng)用12.已知x>0,y>0,lg4x+lg2y=lg8,則12x+4A.3B.94C.4613.已知lg2≈0.3010,lg3≈0.4771,設(shè)N=45×910,則N所在的區(qū)間為()A.(1010,1011)B.(1011,1012)C.(1012,1013)D.(1013,1014)14.(教材習(xí)題改編)已知loga3=m,loga2=n(a>0,且a≠1).(1)求am+2n的值;(2)若0<x<1,x+x-1=a,且m+n=log32+1,求x2-x-2的值.能力提升練題組一對(duì)數(shù)的概念及運(yùn)算1.已知3x=2,log394=y,則x+1A.1B.2C.3D.42.若2a=5,則lg2=()A.aa+1B.aa-13.若lg2=a,lg3=b,則log245=()A.1+aa+3bB.1+4.(多選題)若10a=4,10b=25,5c=4,則下列計(jì)算正確的是()A.a+b=2B.b-a=1C.ab=10D.1a-1c5.化簡823-(log2510)-1+4log26.若實(shí)數(shù)a,b,c滿足3a=4,4b=5,5c=9,則abc=.

7.已知A=-13-20+810.25-(-3)2×823+log53×log325,B=log28.(1)計(jì)算:log25×log154+(lg5)2+lg2×(lg(2)已知a,b均為正實(shí)數(shù),且a,b≠1,若logab+logba=52,ab=ba,求ab題組二對(duì)數(shù)運(yùn)算的綜合應(yīng)用9.假設(shè)在2024年以后,我國每年的GDP(國內(nèi)生產(chǎn)總值)比上一年平均增長8%,那么最有可能實(shí)現(xiàn)GDP翻兩番的目標(biāo)的年份為(參考數(shù)據(jù):lg2≈0.3010,lg3≈0.4771)()A.2032B.2035C.2038D.204210.(多選題)定義(a,b)為a,b之間的一種運(yùn)算,若ac=b,則(a,b)=c,則下列正確的是()A.(4,6)=2×(2,3)B.(2,2)=1C.(2,6)-(2,3)=1D.(4,5)+(4,6)=(4,30)11.已知定義在R上的函數(shù)f(x)滿足f(1-x)=-f(x),f(2-x)=f(x),且當(dāng)x∈0,12時(shí),f(x)=2x,則f(log224)=12.定義在R上的奇函數(shù)f(x)滿足f(1-x)=f(1+x),且f(x)在區(qū)間[1,2]上單調(diào)遞減.若a=f(5),b=f(2-0.513),c=f(log55),則a,b,c按照從小到大的順序排列為13.(1)已知lg2≈0.3010,lg3≈0.4771,求lg45;(2)甲、乙兩人同時(shí)解關(guān)于x的方程:log3x-blogx3+c=0,甲看錯(cuò)了常數(shù)b,得兩根為3和19,乙看錯(cuò)了常數(shù)c,得兩根為127和81,答案與分層梯度式解析4.3對(duì)數(shù)4.3.1對(duì)數(shù)的概念4.3.2對(duì)數(shù)的運(yùn)算基礎(chǔ)過關(guān)練1.B2.BD5.C6.D12.A13.C1.B指對(duì)互化時(shí),底數(shù)不變.故選B.2.BD在選項(xiàng)A中,e3=m化成對(duì)數(shù)式應(yīng)為logem=3,即lnm=3;在選項(xiàng)C中,27-13=13化成對(duì)數(shù)式應(yīng)為log2713=-13;易知B3.答案100解析∵log7[log2(lgx)]=0,∴l(xiāng)og2(lgx)=1,∴l(xiāng)gx=2,故x=100.4.答案17解析∵log32=a,∴3a=2,∴9a=(3a)2=22=4,∴9a+9-a=4+14=175.Clog28=log223=3,因此A錯(cuò)誤;log35+log34=log320≠2,因此B錯(cuò)誤;lg(lg10)=lg1=0,因此C正確;4log29=22log29=6.D2log62+3log633=log6[(2)2×(33)3]=log66=1.故選7.答案10解析eln2+0.125-23+log

39=2+18-28.答案1解析log512-1+7lo9.答案3解析原式=2lg5+23×3lg2+lg5×(1+lg2)+(lg2)2=2(lg5+lg2)+lg5+lg2×lg5+(lg2)=2+1-lg2+lg2×(1-lg2)+(lg2)2=3.10.答案9解析由f(x)=4x,可得f(log23)=4log23=(22)lo易錯(cuò)警示運(yùn)用對(duì)數(shù)恒等式求值時(shí),要將指數(shù)和對(duì)數(shù)的底化為相同的數(shù),再準(zhǔn)確利用公式求值.11.解析(1)由換底公式得log512=lg12lg5=lg3+2lg2lg10?lg2=(2)3log32-23log23×log38+1=2-2log23×log32+log62+log63=2-2+log66=1.12.A∵lg4x+lg2y=lg8,∴4x·2y=8,∴2x+y=3,又x>0,y>0,∴12x+4y=12x+4y當(dāng)且僅當(dāng)y2x=8xy且2x+y=3,即x=12,y=2時(shí)取等號(hào),故12x13.C由于N=45×910,所以lgN=5lg4+10lg9=10lg2+20lg3≈12.552,所以N≈1012.552,故N所在的區(qū)間為(1012,1013).故選C.14.解析(1)由loga3=m,loga2=n得am=3,an=2,因此am+2n=am·a2n=3×22=12.(2)∵m+n=log32+1,∴l(xiāng)oga3+loga2=loga6=log36,即a=3,因此x+x-1=3.于是(x-x-1)2=(x+x-1)2-4=5,由0<x<1知x-x-1<0,從而x-x-1=-5,∴x2-x-2=(x-x-1)(x+x-1)=-35.能力提升練1.A2.C3.D4.AD9.D10.BCD1.A∵log394=y,∴3y=94,因此3y∴3x+y2=3x·3y2=2×32.C由2a=5可得a=log25,則有a=lg5lg2,即alg2=lg5=1-lg2,得(a+1)lg因?yàn)?a=5>1,所以a>0,即a+1≠0,所以lg2=1a+1.故選3.D∵lg2=a,lg3=b,∴由換底公式得log245=lg5lg24=1?lg2lg3+3lg2=1?ab解題模板對(duì)數(shù)式恒等變形的常用策略:一看底,底不同時(shí)用換底公式化為同底;二看真數(shù),利用對(duì)數(shù)的運(yùn)算性質(zhì)將真數(shù)進(jìn)行適當(dāng)變形.解題時(shí)還要考慮到對(duì)數(shù)恒等式及特殊值的應(yīng)用.4.AD若10a=4,10b=25,5c=4,則a=lg4,b=lg25,c=log54,所以a+b=lg4+lg25=lg100=2,選項(xiàng)A正確;b-a=lg25-lg4=lg254≠1,選項(xiàng)B錯(cuò)誤;由基本不等式得ab≤a+b22=1,當(dāng)且僅當(dāng)a=b時(shí)取等號(hào),又a=lg4,b=lg25,所以等號(hào)不成立,即ab<1,選項(xiàng)C錯(cuò)誤;由1a-1c=1lg4-1log54=log410-log45.答案10解析823-(log2510)-1+4=2-lg25+9+1-2lg2=12-lg(25×4)=12-2=10.6.答案2解析因?yàn)?a=4,4b=5,5c=9,所以a=log34,b=log45,c=log59,由換底公式得a=lg4lg3,b=lg5lg4,c=所以abc=lg4lg3×lg5lg4×lg9lg5=lg97.答案-6;2解析A=-13-20+810.25-(-3)2×823+log53×log∵B=log2(4B+2A),∴2B=4B-12,令t=2B(t>0),則t2-t-12=0,解得t=-3(舍去)或t=4,即2B=4,∴B=2.8.解析(1)log25×log154+(lg5)2+lg2×(lg=log25×(-log54)+(lg5)2+lg2×lg5+lg2=log25×(-2log52)+lg5×(lg5+lg2)+lg2=-2+lg5×lg10+lg2=-2+lg5+lg2=-2+lg10=-2+1=-1.(2)令t=logab,則t+1t=52,所以2t2-5t+2=0,即(2t-1)(t-2)=0,所以t=1因此logab=12或logab=2,即a=b2或a2因?yàn)閍b=ba,所以2b=a=b2或b=2a=a2,所以b=2,a=4或a=2,b=4,因此ab=2或ab=9.D設(shè)2024年我國GDP為a(a>0),則n年以后的GDP為a(1+8%)n,若n年以后GDP實(shí)現(xiàn)翻兩番的目標(biāo),則a(1+8%)n=4a,所以n=log1.084=lg4lg1.08=2lg2lg2725=2lg23lg3?2lg5=2lg23lg3?2(1?lg2)=2lg23lg3+2lg2?2≈2×0.30103×0.4771+2×0.3010?2≈18,所以到10.BCD對(duì)于A,設(shè)(4,6)=x1,(2,3)=x2,則4x1=6,2x2=3,得x1=log46=12log26=log26而2×(2,3)=2x2=2log23=log29,故(4,6)=2×(2,3)不成立,因此A錯(cuò)誤;對(duì)于B,設(shè)(2,2)=x3,則2x3=2,得x3=log22=1,因此B對(duì)于C,設(shè)(2,6)=x4,則2x4=6,得x4=log26,由A選項(xiàng)得x2=log則(2,6)-(2,3)=x4-x2=log26-log23=log22=1,因此C正確;對(duì)于D,設(shè)(4,5)=x5,則4x5=5,得x5=log45,由A選項(xiàng)得x1=log設(shè)(4,30)=x6,則4x6=30,得x6=log則x5+x1=log45+log46=log430=x6,即有(4,5)+(4,6)=(4,30),因此D正確.故選BCD.11.答案-4解析由f(x)滿足f(1-x)=-f(x),f(2-x)=f(x),可得f(2-x)=-f(1-x),所以f(x+1)=-f(x),因此f(x+2)=f(x),f(x+4)=f(x),所以f(log224)=f(log224-4)=flog232=-f因?yàn)?=log21<log243<log22=12,所以flog243=2log24312.答案c,b,a解析由f(1-x)=f(1+x)知函數(shù)f(x)的圖象關(guān)于直線x=1對(duì)稱,所以c=f(log55)=f12=f3由f(1-x)=f(1+x)知f(2-x)=f(x),由f(x)是R上的奇函數(shù)知f(x)=-f(-x),所以f(2-x)=-f(-x),所以f(2+x)=-f(x),所以f(4+x)=-f(2+x)=f(x),則a=f(5)=f(1),又12<0.513<1,所