新版高中数学人教A版必修1习题第二章基本初等函数（Ⅰ）2.2.1.1

2.2对数函数2.2.1对数与对数运算第1课时对数课时过关·能力提升基础巩固1.log5b=2化为指数式是()A.5b=2 B.b5=2 C.52=b D.b2=5答案:C2.3b=5化为对数式是()A.logb3=5 B.log35=bC.log5b=3 D.log53=b答案:B3.对数lne2的值为()A.2 B.e C.10 D.e2答案:A4.已知logx8=3,则x的值为()A.12 B.2C.3D.解析:由题意,得x3=8=23,即x=2.答案:B5.若loga7b=c(a>0,且a≠1,b>0),则有(A.b=a7c B.b7=ac C.b=7ac D.b=c7a解析:∵loga7b=c,∴ac=7∴(ac)7=(7b)7.∴a7c=b答案:A6.log23278解析:设log23278=m又278=32∴m=3,即log2327答案:37.已知log2x=3,则x-13=解析:∵log2x=3,∴x=23=8,∴x-13=8-13=(2答案:18.若log31-2x9=0,则解析:由题意,得1-2x9=1,答案:49.21+log23+log31解析:令log3127=x,∴3x=127=33,∴x=∴原式=2·2log23+log3127=2×答案:310.求值:(32)6+log33+log1解:设log1216=t,则12t=16,∴2t∴t=4.∴原式=263+14=11.求下列各式中的x.(1)logx27=32;(2)log2x=2(3)logx(3+22)=2;(4)log5(log2x)=0;(5)x=log2719解:(1)由logx27=32,得x32=27,故x=2723=(2)由log2x=23,得2-23=x,(3)由logx(3+22)=2,得3+22=x2,即x=(3+22)-(4)由log5(log2x)=0,得log2x=1,故x=21=2.(5)由x=log2719,得27x=19,即33x=3故x=23能力提升1.若7x=8,则x等于()A.87 B.log87 C.log78 D.log7答案:C2.23=18化为对数式为(A.log182=3 B.log18C.log218=3 D.log2(3)=答案:C3.在log(x2)(5x)中,实数x的取值范围是()A.(∞,2)∪(5,+∞) B.(2,3)∪(3,5)C.(2,5) D.(3,4)解析:由5-x>0,x-2>0答案:B4.以下结论正确的个数是()①lg(lg10)=0; ②lg(lne)=0;③若10=lgx,则x=10; ④若e=lnx,则x=e2.A.1 B.2 C.3 D.4解析:①中,∵lg(lg10)=lg1=0,∴①正确;②中,∵lg(lne)=lg1=0,∴②正确;③中,∵10=lgx,∴x=1010,∴③不正确;④中,∵e=lnx,∴x=ee,∴④不正确.答案:B5.若a=lg2,b=lg3,则100a-b2解析:∵a=lg2,∴10a=2.∵b=lg3,∴10b=3.∴100a答案:46.方程9x6·3x7=0的解是x=.

解析:设3x=t(t>0),则t26t7=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:2★8.求下列各式的值:(1)log2(log3(log464));(2)log2.56.253lo解:(1)∵43=64,∴log464=3,∴log2(log3(log464))=log2(