3.2.1对数及其运算(3)A级基础巩固一、选择题1.若log5·log36·log6x=2,则x=(D)A.9B.C.25D.[解析]∵log5·log36·log6x=2,∴··=2,∴lgx=-2lg5=lg5-2,∴x=.2.+等于(C)A.lg3B.-lg3C.D.-[解析]+=+=+=+==.3.eln3-e-ln2等于(C)A.1B.2C.D.3[解析]eln3-e-ln2=eloge3-=3-=.4.已知log32=a,3b=5,则log3用a、b表示为(A)A.(a+b+1)B.(a+b)+1C.(a+b+1)D.a+b+1[解析]∵3b=5,∴b=log35.log3=log330=(log33+log32+log35)=(1+a+b),选A.二、填空题5.计算log43·log98=.[解析]log43·log98=·=·=.6.已知f(3x)=2x·log23,则f(21007)的值等于__2_014__.[解析]令3x=t,∴x=log3t,∴f(t)=2log3t·log23=2··=2log2t,∴f(21007)=2log221007=2×1007=2014.三、解答题7.若log37·log29·log49m=log4,求m的值.[解析]∵log37·log29·log49m=log4,∴··==-,∴lgm=-lg2=lg2-,∴m=2-=.8.计算3log34-27-lg0.01+lne3的值.[解析]3log34-27-lg0.01+lne3=4-32-lg10-2+lne3=4-9+2+3=0.B级素养提升一、选择题1.设2a=5b=m,且+=2,则m=(A)A.B.10C.20D.100[解析]∵2a=5b=m,∴a=log2m,b=log5m,∴+=+=logm2+logm5=logm10=2,∴m=.故选A.2.方程eln|x|=2的解是(C)A.-2B.2C.-2或2D.4[解析]∵eln|x|=2,∴|x|=2,∴x=-2或2.二、填空题3.=__1__.[解析]===1.4.若mlog35=1,n=5m,则n的值为__3__.[解析]∵mlog35=1,∴m==log53.∴n=5m=5log53=3.三、解答题5.已知log98=p,log2725=q,试用p、q表示log52.[解析]∵p=log98=log32,q=log2725=log35,∴log52===.C级能力拔高1.已知x、y、z均大于1,a≠0,logza=24,logya=40,log(xyz)a=12,求logxa.[解析]由logza=24得logaz=,由logya=40得logay=,由log(xyz)a=12得loga(xyz)=,即logax+logay+logaz=.∴logax++=,解得logax=,∴logxa=60.2.已知logax+3logxa-logxy=3(a>1).(1)若设x=at,试用a、t表示y;(2)若当0<t≤2时,y有最小值8,求a和x的值.[解析](1)由换底公式,得logax+-=3(a>1),∴logay=(logax)2-3logax+3,当x=at时,logax=logaat=t,∴logay=t2-3t+3,故y=at2-3t+3(t≠0).(2)y=a(t-)2+,∵0<t≤2,a>1,∴当t=时,ymin=a=8,∴a=16,此时x=a=64.