對(duì)數(shù)的概念+練習(xí) 高一上學(xué)期數(shù)學(xué)北師大版（2019）必修第一冊(cè)

第四章對(duì)數(shù)運(yùn)算與對(duì)數(shù)函數(shù)§1對(duì)數(shù)的概念一、選擇題(本大題共9小題,每小題5分,共45分)1.[2024·黑龍江哈爾濱九中高一期中]將23=8化為對(duì)數(shù)式,則正確的對(duì)數(shù)式是 ()A.log23=8 B.log28=3C.log82=3 D.log32=82.已知log3a=2,則a= ()A.6 B.7C.8 D.93.[2024·貴州貴陽(yáng)清鎮(zhèn)博雅實(shí)驗(yàn)學(xué)校高一月考]使式子log(3x-1)(2-x)有意義的x的取值范圍是 ()A.x>2B.13<x<C.13<x<2且x≠D.x<24.下列結(jié)論中,正確的是 ()A.ln(lne)=1B.lg(lg10)=10C.若log4x=-2,則x=1D.若3x=8,則x=log385.若logx7y=z,其中x>0且x≠1,y>0,則 (A.y7=xz B.y=x7zC.y=7xz D.y=z7x6.已知log7[log3(log2x)]=0,那么x-12= A.13 B.C.33 D.7.已知正實(shí)數(shù)a,b滿足ab=ba,且logab=2,a≠1,則ab= ()A.12 B.2 C.4 D.8.(多選題)給出下列四個(gè)命題:①lg10=1;②若2x=N,則x=log2N;③lg(lne)=1;④ln(lg10)=1.其中真命題是 ()A.① B.② C.③ D.④9.(多選題)已知a,b均為不等于1的正實(shí)數(shù),若logab+1logab=52,則a,bA.a=b2 B.a=5b2C.5a2=b D.a2=b二、填空題(本大題共3小題,每小題5分,共15分)10.已知f(10x)=x,則f(3)=.

11.[2024·寧夏吳忠秦寧中學(xué)高一月考]若log2(log4X)=log3(log2Y)=1,則X+Y的值為.

12.方程2(log4x)2-log4x=6的解是.

三、解答題(本大題共2小題,共20分)13.(10分)互化下列的指數(shù)式與對(duì)數(shù)式:(1)33=27;(2)4-12(3)ln3e=13;(4)lg0.1=-14.(10分)(1)求下列各式中的x的值:①log13x=-3;②logx64=4(x>0且x≠1);③lg0.(2)計(jì)算:2lne+lg1+3log32-15.(5分)已知f(ex)=xlg5,則f(1)+f(e)=.

16.(15分)若log4{2log2[1+log2(1+log2x)]}=12,求x的值第四章對(duì)數(shù)運(yùn)算與對(duì)數(shù)函數(shù)§1對(duì)數(shù)的概念1.B[解析]23=8化為對(duì)數(shù)式得log28=3,故選B.2.D[解析]由log3a=2,得a=32=9,故選D.3.C[解析]由式子log(3x-1)(2-x)有意義,得3x-1>0,3x-1≠1,2-x4.D[解析]ln(lne)=ln1=0,A錯(cuò)誤;lg(lg10)=lg1=0,B錯(cuò)誤;若log4x=-2,則x=4-2=116,C錯(cuò)誤;若3x=8,則x=log38,D正確.故選D5.B[解析]由對(duì)數(shù)的定義,將已知的對(duì)數(shù)式轉(zhuǎn)化為指數(shù)式,可得y=x7z.6.D[解析]∵log7[log3(log2x)]=0,∴l(xiāng)og3(log2x)=1,∴l(xiāng)og2x=3,∴x=8,∴x-12=18=27.D[解析]∵正實(shí)數(shù)a,b滿足logab=2,a≠1,∴a2=b,又ab=ba,∴aa2=a2a,即a2=2a,解得a=2或a=0(舍去),∴b=4,故ab=8,故選8.AB[解析]lg10=log1010=1,①是真命題;根據(jù)指數(shù)式和對(duì)數(shù)式的互化可知②是真命題;lg(lne)=lg1=0,③是假命題;ln(lg10)=0,④是假命題.故選AB.9.AD[解析]令logab=x,則logab+1logab=52可化為x+1x=52,解得x=2或x=12.當(dāng)logab=12時(shí),得a=b2,A正確;當(dāng)logab=2時(shí),可得10.lg3[解析]令10x=3,則x=lg3,所以f(3)=lg3.11.24[解析]因?yàn)閘og2(log4X)=log3(log2Y)=1,所以log4X=2,log2Y=3,所以X=42=16,Y=23=8,所以X+Y=16+8=24.12.x=18或x=16[解析]令log4x=t,則原方程化為2t2-t-6=0,得(2t+3)(t-2)=0,∴t=2或t=-32,即log4x=2或log4x=-32,解得x=16或x=18,∴方程2(log4x)2-log4x=6的解是x=113.解:由對(duì)數(shù)的定義,得(1)log327=3.(2)log412=-1(3)e13=(4)10-1=0.1.14.解:(1)①由log13x=-3可得x=13-3=②由logx64=4可得x4=64,又x>0且x≠1,所以x=22.③由lg0.00001=x得lg10-5=x,即10x=10-5,所以x=-5.(2)2lne+lg1+3log32-log216=21+2-15.lg5[解析]方法一:令t=ex,則x=lnt,所以f(t)=lnt·lg5,即f(x)=lnx·lg5,所以f(1)+f(e)=ln1×lg5+lne×lg5=0×lg5+1×lg5=lg5.方法二:令ex=1,則x=0,令ex=e,則x=1,所以f(1)+f(e)=0×lg5+1×lg5=lg5.16.解:由log4{2log2[1+log2(1+l