高中數(shù)學(xué)第二章對(duì)數(shù)函數(shù)2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算第1課時(shí)對(duì)數(shù)練習(xí)新人教A版.docx

2.2對(duì)數(shù)函數(shù)2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算第1課時(shí)對(duì)數(shù)課時(shí)過關(guān)能力提升基礎(chǔ)鞏固1.log5b=2化為指數(shù)式是()A.5b=2B.b5=2C.52=bD.b2=5答案:C2.3b=5化為對(duì)數(shù)式是()A.logb3=5B.log35=bC.log5b=3D.log53=b答案:B3.對(duì)數(shù)ln e2的值為()A.2B.eC.10D.e2答案:A4.已知logx8=3,則x的值為()A.12B.2C.3D.4解析:由題意,得x3=8=23,即x=2.答案:B5.若loga7b=c(a0,且a1,b0),則有()A.b=a7cB.b7=acC.b=7acD.b=c7a解析:loga7b=c,ac=7b.(ac)7=(7b)7.a7c=b.答案:A6.log23278=.解析:設(shè)log23278=m,則23m=278.又278=323=23-3,23m=23-3,m=-3,即log23278=-3.答案:-37.已知log2x=3,則x-13=.解析:log2x=3,x=23=8,x-13=8-13=(23)-13=2-1=12.答案:128.若log31-2x9=0,則x=.解析:由題意,得1-2x9=1,解得x=-4.答案:-49.21+log23+log3127=.解析:令log3127=x,3x=127=3-3,x=-3.原式=22log23+log3127=23-3=3.答案:310.求值:(32)6+log33+log1216.解:設(shè)log1216=t,則12t=16,2-t=24,t=-4.原式=263+1-4=1.11.求下列各式中的x.(1)logx27=32;(2)log2x=-23;(3)logx(3+22)=-2;(4)log5(log2x)=0;(5)x=log2719.解:(1)由logx27=32,得x32=27,故x=2723=32=9.(2)由log2x=-23,得2-23=x,故x=1322=322.(3)由logx(3+22)=-2,得3+22=x-2,即x=(3+22)-12=2-1.(4)由log5(log2x)=0,得log2x=1,故x=21=2.(5)由x=log2719,得27x=19,即33x=3-2,故x=-23.能力提升1.若7x=8,則x等于()A.87B.log87C.log78D.log7x答案:C2.2-3=18化為對(duì)數(shù)式為()A.log182=-3B.log18(-3)=2C.log218=-3D.log2(-3)=18答案:C3.在log(x-2)(5-x)中,實(shí)數(shù)x的取值范圍是()A.(-,2)(5,+)B.(2,3)(3,5)C.(2,5)D.(3,4)解析:由5-x0,x-20,x-21,得2x3或3x0),則t2-6t-7=0,解得t=7或t=-1(舍去),3x=7.x=log37.答案:log377.方程log2(log3(x+1)=0的解x=.解析:log2(log3(x+1)=0,log3(x+1)=1,x+1=3,解得x=2.答案:28.求下列各式的值:(1)log2(log3(log464);(2)log2.56.25-3log32+14-12.解:(1)43=64,log464=3,log2(log3(log464)=log2(log33)=log21=0.(2)令log2.