2019_2020学年高中数学第二章基本初等函数（Ⅰ）2.2.1指数函数及其性质（第1课时）对数练习新人教A版.docx

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=2B.b5=2C.52=bD.b2=5答案:C2.3b=5化为对数式是()A.logb3=5B.log35=bC.log5b=3D.log53=b答案:B3.使对数loga(-2a+1)有意义的a的取值范围为()A.a12,且a1B.0a0,且a1D.a0,a1,-2a+10,解得0a0,且a1,b0),则有()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:设log23278=m,则23m=278.又278=323=23-3,23m=23-3,m=-3,即log23278=-3.答案:-37.已知log2x=3,则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,则x=_.解析:由题意,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=-23.11.若log12x=m,log14y=m+2,求x2y的值.解:log12x=m,12m=x,x2=122m.log14y=m+2,14m+2=y,y=122m+4.x2y=122m122m+4=122m-(2m+4)=12-4=16.能力提升1.12-1+log0.54的值为()A.6B.72C.8D.37解析:12-1+log0.54=12-112log124=24=8.答案:C2.已知log7(log3(log2x)=0,那么x-12等于()A.13 B.36 C.24 D.33解析:由条件,知log3(log2x)=1,所以log2x=3,即x=23=8,所以x-12=8-12=1812=122=24.答案:C3.在log(x-2)(5-x)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-x0,x-20,x-21,得2x3或3x0),则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.(1)已知log189=a,log1854=b,求182a-b的值;(2)已知logx27=31+log32,求x的值.解:(1)log189=a,log1854=b,18a=9,18b=54