课时作业19对数的运算时间:45分钟分值:100分一、选择题(每小题6分,共计36分)1.若lgx=lga+2lgb-3lgc,则x等于()A.a+2b-3cB.C.D.ab2-c3解析:lgx=lga+2lgb-3lgc=lg,∴x=.答案:C2.化简:log2+log2+log2+…+log2等于()A.5B.4C.-5D.-4解析:原式=log2(×××…×)=log2=-5.答案:C3.若lnx-lny=a,则ln3-ln3=()A.B.aC.D.3a解析:ln3-ln3=3=3(lnx-ln2-lny+ln2)=3(lnx-lny)=3a.答案:D4.设log34·log48·log8m=log416,则m的值为()A.B.9C.18D.27解析:由题意得··=log416=log442=2,∴=2,即lgm=2lg3=lg9.∴m=9,选B.答案:B5.定义新运算“&”与“*”:x&y=xy-1,x*y=log(x-1)y,则函数f(x)=是()A.奇函数B.偶函数C.非奇非偶函数D.既是奇函数又是偶函数答案:A16.已知2x=3,log4=y,则x+2y等于()A.3B.8C.4D.log48解析:∵2x=3,∴x=log23.又log4=y,∴x+2y=log23+2log4=log23+2(log48-log43)=log23+2=log23+3-log23=3.故选A.答案:A二、填空题(每小题8分,共计24分)7.|1+lg0.001|++lg6-lg0.03=________.解析:原式=|1+lg10-3|++lg6-lg=|1-3|++lg6-lg3+2=2+2-lg2+lg6-lg3+2=6+lg=6.答案:68.(lg5)2+2lg2-(lg2)2+log23·log34=________.解析:原式=(lg5)2-(lg2)2+2lg2+log24=(lg5+lg2)(lg5-lg2)+2lg2+2=lg5-lg2+2lg2+2=lg5+lg2+2=3.答案:39.若lga,lgb是方程2x2-4x+1=0的两个根,则(lg)2=________.解析:由韦达定理,得lga+lgb=2,lga·lgb=,则(lg)2=(lga-lgb)2=(lga+lgb)2-4lga·lgb=22-4×=2.答案:2三、解答题(共计40分)10.(10分)计算:(1)log535-2log5+log57-log51.8;(2)log2+log212-log242-1.解:(1)原式=log5(5×7)-2(log57-log53)+log57-log5=log55+log57-2log57+2log53+log57-2log53+log55=2.(2)原式=log2+log212-log2-log22=log2=log2==-.11.(15分)计算:(1)(log32+log92)·(log43+log83);(2).解:(1)原式=·=·2=·=;(2)分子=lg5(3+3lg2)+3(lg2)2=3lg5+3lg2(lg5+lg2)=3,分母=(lg6+2)-lg=lg6+2-lg=4,∴原式=.——能力提升——12.(15分)已知100m=5,10n=2.(1)求2m+n的值;(2)x1、x2、…、x10均为正实数,若函数f(x)=logax(a>0且a≠1),且f(x1·x2·…·x10)=2m+n,求f(x)+f(x)+…+f(x)的值.解:(1)法一∵100m=102m=5,∴102m·10n=102m+n=10,∴2m+n=1.法二∵100m=5,∴2m=lg5∵10n=2,∴n=lg2,∴2m+n=lg5+lg2=lg10=1.(2)由对数的运算性质知loga(x1·x2…x10)=logax1+logax2+…+logax10,logax2=2logax且由(1)知2m+n=1,∴f(x1x2…x10)=f(x1)+f(x2)+…+f(x10)=1,∴f(x)+f(x)+…+f(x)=2[f(x1)+f(x2)+…+f(x10)]=2×1=2.3