1、高考资源网() 您身边的高考专家第三章3.23.2.1第3课时一、选择题1log52log425等于()A1BC1D2答案C解析log52log4251.2化简logbloga的值为()A0 B1C2logab D2logab答案A解析logbloga0.3(20132014学年度河北衡水中学高一期中测试)若xlog341,则4x4x的值为()A. BC2 D1答案B解析xlog341,xlog43,4x4log433,4x,4x4x3.4若log5log36log6x2,则x()A9 BC25 D答案D解析log5log36log6x2,2,lgx2lg5lg52,x.5.等于()Alg3
2、Blg3 C. D答案C解析.6eln3eln2等于()A1 B2 C. D3答案C解析eln3eln2eloge33.二、填空题7(20132014学年度广东湛江一中高一上学期期中测试)计算log43log98_.答案解析log43log98.8已知f(3x)2xlog23,则f(21 005)的值等于_答案2 010解析令3xt,xlog3t,f(t)2log3tlog2322log2t,f(21 005)2log221 00521 0052 010.三、解答题9若log37log29log49mlog4,求m的值解析log37log29log49mlog4,lgmlg2lg2,m2.一、
3、选择题1已知log32a,3b5,则log3用a、b表示为()A.(ab1) B(ab)1C.(ab1) Dab1答案A解析3b5,blog35.log3log330(log33log32log35)(1ab),选A.2已知logax2,logbx1,logcx4,则logx(abc)等于()A. B C. D答案D解析由题意得xa2,xb,xc4,(abc)4x7,abcx,logx(abc).3设2a5bm,且2,则m()A. B10C20 D100答案A解析2a5bm,alog2m,blog5m,logm2logm5logm102,m.故选A.4方程eln|x|2的解是()A2 B2C2
4、或2 D4答案C解析eln|x|2,|x|2,x2或2.二、填空题5._.答案1解析1.6若mlog351,n5m,则n的值为_答案3解析mlog351,mlog53.n5m5log533.三、解答题7已知log98p,log2725q,试用p、q表示log52.解析plog98log32,qlog2725log35,log52.8已知x,y,z均大于1,a0,logza24,logya40,log(xyz)a12,求logxa.解析由logza24得logaz,由logya40得logay,由log(xyz)a12得loga(xyz),即logaxlogaylogaz.logax,解得logax,logxa60.9已知logax3logxalogxy3(a1)(1)若设xat,试用a,t表示y;(2)若当0t2时,y有最小值8,求a和x的值解析(1)由换底公式,得logax3(a1),logay(logax)23logax3,当xat时,logaxlogaatt,logayt23t3,故yat23t3(t0)(2)ya(t)2,0t2,a1,当t时,ymina8,a16,此时xa64.- 5 - 版权所有高考资源网