2024-2025學(xué)年高中數(shù)學(xué)第二章基本初等函數(shù)Ⅰ2.2.1指數(shù)函數(shù)及其性質(zhì)第1課時對數(shù)練習(xí)含解析新人教A版必修1

PAGEPAGE5第1課時對數(shù)課時過關(guān)·實(shí)力提升基礎(chǔ)鞏固1.log5b=2化為指數(shù)式是()A.5b=2 B.b5=2 C.52=b D.b2=5答案:C2.3b=5化為對數(shù)式是()A.logb3=5 B.log35=bC.log5b=3 D.log53=b答案:B3.使對數(shù)loga(-2a+1)有意義的a的取值范圍為()A.a>12,且a≠1 BC.a>0,且a≠1 D.a<解析:由對數(shù)的概念可知使對數(shù)loga(-2a+1)有意義的a需滿足a>0答案:B4.已知logx8=3,則x的值為()A.BCD解析:由題意,得x3=8=23,即x=2.答案:B5.若loga7b=c(a>0,且aA.b=a7c B.b7=ac C.b=7ac D.b=c7a解析:∵loga∴(ac)7=(答案:A6.log解析:設(shè)lo又∴m=-3,即lo答案:-37.已知log2x=3,則x解析:∵log2x=3,∴x=23=8,∴答案:18.若log3解析:由題意,得1-2答案:-49.21+lo解析:令log∴原式=2·2答案:310.求下列各式中的x.(1)logx27=(3)logx(3+2(5)x=log27解:(1)由logx27=32(2)由log2x=-23(3)由logx(3+22)=-即x=(3+2(4)由log5(log2x)=0,得log2x=1,故x=21=2.(5)由x=log2719,得27x=19,即33故x=-11.若log解:∵lo∵lo∴實(shí)力提升1.12A.6 B.解析:1答案:C2.已知log7(log3(log2x))=0,那么xA.解析:由條件,知log3(log2x)=1,所以log2x=3,即x=23=8,所以答案: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-x>0,答案:B4.以下結(jié)論正確的個數(shù)是()①lg(lg10)=0; ②lg(lne)=0;③若10=lgx,則x=10; ④若e=lnx,則x=e2.A.1 B.2 C.3 D.4解析:①中,∵lg(lg10)=lg1=0,∴①正確;②中,∵lg(lne)=lg1=0,∴②正確;③中,∵10=lgx,∴x=1010,∴③不正確;④中,∵e=lnx,∴x=ee,∴④不正確.答案:B5.若a=lg2,b=lg3,則100解析:∵a=lg2,∴10a=2.∵b=lg3,∴10b=3.∴10答案:46.方程9x-6·3x-7=0的解是x=.

解析:設(shè)3x=t(t>0),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,∴l(xiāng)og3(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=解:(1)∵log1