高一数学必修1(北师大版)同步练习3-4
3-4 对数 基 础 巩 固一、选择题1lg20lg50的值为()A70 B1000 C3 D.答案C解析lg20lg50lg10003.故选C.2已知alog32,那么log382log36用a表示是()Aa2 B5a2C3a(1a)2 D3aa21答案A解析log382log36log3232(log32log33)3log322(log321)3a2(a1)a2.故选A.3若a>0,a1,x>0,nN,则下列各式:(logax)nnlogax;(logax)nlogaxn;logaxloga;logax;loga;logaxnnlogax.其中成立的有()A3个 B4个 C5个 D6个答案A解析正确;错误4若log2log3(log4x)log3log4(log2y)log4log2(log3z)0,则xyz()A50 B58 C89 D111答案C解析log2log3(log4x)0,log3(log4x)1,log4x3,x4364,同理y16,z9,xyz89,故选C.5如果f(10x)x,则f(3)等于()Alog310 Blg3C103 D310答案B解析令10x3,xlg3.故选B.6方程log3(x1)log9(x5)的解为()Ax1 Bx1或x4Cx4 Dx1且x4答案C解析一定要注意对数的真数大于零,即,解得x4,选C.二、填空题7求值:答案(1)25(2)(3)72解析 8若正数m,满足10m1<2512<10m,则m_.(lg20.3010)答案155解析10m1<2512<10m,m1<512lg2<m,154.112<m<155.112,又mN,m155.三、解答题9计算:(log2125log425log85)·(log52log254log1258)解析解法一:原式(log253)(log52)log25·(3log52)13log25·13.解法二:原式13.能 力 提 升一、选择题1. Alg3 Blg3C. D答案C解析.2设a、b、c均为正实数,且3a4b6c,则有()A. B.C. D.答案B解析令3a4b6ct,alog3t,blog4t,clog6t,故选B.二、填空题3(2011·陕西文)设f(x)则ff(2)_.答案2解析本题考查分段函数求值方法是“由里向外”层层去掉“f”f(2)102,ff(2)f(102)lg1022.4若mlog351,n5m5m,则n的值为_答案解析mlog53.3.三、解答题5若a、b是方程2lg2xlgx410的两个实数根,求lg(ab)(logablogba)的值解析原方程可化为2lg2x4lgx10.依题意知,lgalgb2,lga·lgb,lg(ab)(logablogba)(lgalgb)2×12.6已知lgxlgy2lg(x2y),求log的值解析由已知得lg(xy)lg(x2y)2,从而有xy(x2y)2整理得x25xy4y20,即(xy)(x4y)0,xy或x4y.但由x>0,y>0,x2y>0得x>2y>0.xy应舍去,故4.loglog44.7已知二次函数f(x)(lga)x22x4lga的最大值为3,求a的值解析原函数式可化成f(x)lga(x)24lga.由已知,f(x)有最大值3,所以lga<0,并且4lga3,整理得4(lga)23lga10,解得lga1,或lga.又lga<0,lga,a10.