2019-2020学年新教材高中数学 课时素养评价五 对数运算法则 新人教B版必修2.doc

课时素养评价 五对数运算法则 (25分钟50分)一、选择题(每小题4分,共16分.多选题全部选对的得4分,选对但不全的得2分,有选错的得0分)1.(多选题)已知x,y为正实数,则()A.2ln x+ln y=2ln x+2ln yB.2ln(x+y)=2ln x2ln yC.2ln xln y=(2ln x)ln yD.2ln(xy)=2ln x2ln y【解析】选C、D.根据指数与对数的运算性质可得2ln xln y=(2ln x)ln y,2ln(xy)=2ln x+ln y=2ln x2ln y,可知:C,D正确,而A,B都不正确.2.(2019温州高二检测)lg-11002=()A.-4B.4C.10D.-10【解析】选A.lg-11002=lg 10-4=-4.3.若lg x=m,lg y=n,则lgx-lgy102的值为()A.12m-2n-2B.12m-2n-1C.12m-2n+1 D.12m-2n+2【解析】选D.因为lg x=m,lg y=n,所以lg x-lgy102=12lg x-2lg y+2=12m-2n+2.4.(2019泸州高二检测)实数a,b满足2a=5b=10,则下列关系正确的是()A.2a+1b=2B.1a+1b=1C.1a+2b=2D.1a+2b=12【解析】选B.因为2a=5b=10,所以a=log2 10,b=log5 10,所以1a=lg 2,1b=lg 5,所以1a+1b=lg 2+lg 5=lg(25)=1.二、填空题(每小题4分,共8分)5.(2019姜堰高二检测)计算log525+lne-(0.64)12=_.【解析】原式=2+12-45=1710.答案:17106.(2019大庆高一检测)已知x0,y0,若2x8y=16,则x+3y=_,则2-1+log2x+log927y=_.【解析】根据题意,若2x8y=16,则2x+3y=24,则x+3y=4,则2-1+log2x+log927y=x2+3y2=12(x+3y)=2.答案:42三、解答题(共26分)7.(12分)(1)计算:(lg 2)2+(lg 2+3)lg 5+lg 4.(2)已知log53=a,log54=b,用a,b表示log25144.【解析】(1)原式=(lg 2)2+lg 2lg 5+3lg 5+lg 4=lg 2(lg 2+lg 5)+lg 5+2(lg 2+lg 5)=lg 2+lg 5+2=3.(2)因为log53=a,log54=b,所以log25144=log512=log53+log54=a+b.8.(14分)(2019银川高一检测)求下列各式的值:(1)log327+lg 25+lg4+7log712+(-9.8)0.(2)lg 25+23lg 8+lg 5lg 20+(lg 2)2.【解析】(1)log327+lg 25+lg4+7log712+(-9.8)0=32+2+12+1=5.(2)lg 25+23lg 8+lg 5lg 20+(lg 2)2=2lg 5+2lg 2+lg 5(2lg 2+lg 5)+(lg 2)2=2+lg 2lg 5+lg 5+(lg 2)2=2+lg 2(lg 2+lg 5)+lg 5=2+lg 2+lg 5=3. (15分钟30分)1.(4分)(2019台州高一检测)已知实数a,b满足ab=ba,且logab=2,则ab=()A.12B.2C.4D.8【解析】选D.因为实数a,b满足logab=2,故a2=b,又由ab=ba得aa2=a2a,解得:a=2,或a=0(舍去),故b=4,ab=8.2.(4分)(2019碑林高二检测)已知29x-28=13-x,则x=()A.log 37-log32B.log134C.2log32D.log37【解析】选C.已知29x-28=13-x,所以2(3x)2-28-3x=0,即(3x-4)(23x+7)=0,解得3x=4,则x=log34=2log32.3.(4分)lg252+3lg 2-12-1+e0=_.【解析】lg252+3lg 2-12-1+e0=lg2528-2+1=1.答案:1【加练固】(-4)2+log2(4725)-ln1e=_.【解析】(-4)2+log2(4725)-ln1e=4-+log2219+=4+19=23.答案:234.(4分)(2019宁波高一检测)已知logab+3logba=132,则logab=_,当ab1时,a+b4a2+b2的值为_.【解析】因为logab+3logba=132,所以logab+3logab=132,所以2(logab)2-13logab+6=0,解得logab=6或12,因为ab1,所以0logab0),所以x=log2k,y=log3k,z=log5k,所以1x=logk2,1y=logk3,1z=logk5,由1x+1y+1z=1,得logk2+logk3+logk5=logk30=1,所以k=30,所以x=log230=1+log215,y=log330=1+log310,z=log530=1+log56.1.方程log2(x+2)=1+log4(6-x)的解为x=_.【解析】由log2(x+2)=1+log4(6-x),得:log2(x+2)=log2(26-x),所以x+2=26-x,解得x=2或-10(舍去).答案:22.若a,b,cN*,且满足a2+b2=c2.(1)求log21+b+ca+log21+a-cb的值.(2)若log41+b+ca=1,log8(a+b-c)=23,求a,b,c的值.【解析】(1)因为a2+b2=c2,所以log21+b+ca+log21+a-cb=log21+b+ca1+a-cb=log2(a+b+c)(a+b-c)ab=log2a2+b2-c2+