1、4.3.2对数的运算课后篇巩固提升合格考达标练1.2log510+log50.25=() A.0B.1C.2D.4答案C解析原式=log5102+log50.25=log5(1000.25)=log525=2.2.(2021河南郑州高一期末)已知alog32=1,则2a=()A.13B.1C.2D.3答案D解析alog32=1=log32a,故2a=3.故选D.3.(2021吉林公主岭高一期末)log28+lg 25+lg 4+6log612+9.80=()A.1B.4C.5D.7答案C解析原式=32log22+lg(254)+12+1=2+2+1=5.故选C.4.(多选题)(2021江苏连云
2、港高一期末)若x0,y0,n0,mR,则下列各式中,恒等的是()A.lg x+lg y=lg(x+y)B.lgxy=lg x-lg yC.logxnym=mnlogxyD.lgx1n=lgxn答案BCD解析因为x0,y0,n0,mR,则lg x+lg y=lg(xy),故A错误;lgxy=lg x-lg y,故B正确;logxnym=mnlogxy,故C正确;lg x1n=lgxn,故D正确.故选BCD.5.若2lg(x-2y)=lg x+lg y(x2y0),则yx的值为()A.4B.1或14C.1或4D.14答案D解析2lg(x-2y)=lg x+lg y(x2y0),lg(x-2y)2=
3、lg xy,(x-2y)2=xy,x2-5xy+4y2=0,(x-y)(x-4y)=0,x=y或x=4y.x-2y0,且x0,y0,xy,yx=14.6.计算:2713+lg 4+2lg 5-eln 3=.答案2解析由题意得2713+lg 4+2lg 5-eln 3=(33)13+(lg 4+lg 25)-eln 3=3+2-3=2.7.log35log46log57log68log79=.答案3解析log35log46log57log68log79=lg5lg3lg6lg4lg7lg5lg8lg6lg9lg7=lg8lg9lg3lg4=3lg22lg3lg32lg2=3.8.计算:(1)lg
4、2+lg5-lg8lg50-lg40;(2)lg12-lg58+lg54-log92log43.解(1)原式=lg258lg5040=lg54lg54=1.(2)(方法一)原式=lg1258+lg54-lg2lg9lg3lg4=lg4554-lg22lg3lg32lg2=lg 1-14=-14.(方法二)原式=(lg 1-lg 2)-(lg 5-lg 8)+(lg 5-lg 4)-lg2lg9lg3lg4=-lg 2+lg 8-lg 4-lg22lg3lg32lg2=-(lg 2+lg 4)+lg 8-14=-lg(24)+lg 8-14=-14.等级考提升练9.(2021北京昌平高一期末)已
5、知2x=3,log289=y,则2x+y=()A.3B.4C.8D.9答案A解析由2x=3可知x=log23,且y=log289.2x+y=2log23+log289=log23289=log28=3.10.(2021浙江嘉兴高一期末)设lg 3=a,lg 5=b,则log212的值为()A.2b-a+21-bB.2b-a+2b-1C.a-2b+21-bD.a-2b+21+b答案C解析根据换底公式和对数运算性质得log212=lg12lg2=lg3+2lg2lg2=lg3+2lg105lg105=lg3+2-2lg51-lg5=2+a-2b1-b.故选C.11.(多选题)设a,b,c都是正数,
6、且4a=6b=9c,那么()A.ab+bc=2acB.ab+bc=acC.2c=2a+1bD.1c=2b-1a答案AD解析由题意,设4a=6b=9c=k(k0),则a=log4k,b=log6k,c=log9k,对于选项A,由ab+bc=2ac,可得bc+ba=2,因为bc+ba=log6klog9k+log6klog4k=logk9logk6+logk4logk6=log69+log64=log636=2,故A正确,B错误;对于选项C,2a+1b=2log4k+1log6k=2logk4+logk6=logk96,2c=2log9k=2logk9=logk81,故2c2a+1b,故C错误;对
7、于选项D,2b-1a=2log6k-1log4k=2logk6-logk4=logk9,1c=1log9k=logk9,故1c=2b-1a,故D正确.12.已知ab1,若logab+logba=52,ab=ba,则a=,b=.答案42解析logab+logba=logab+1logab=52,logab=2或logab=12.ab1,logab0,2x-11,5x2+3x-170,5x2+3x-17=(2x-1)2,即2x-10,2x-11,5x2+3x-17=4x2-4x+1,解得x=2或x=-9(舍).(2)由logx4+log2x=3(x0,且x1),得2logx2+log2x-3=0,
8、令log2x=t,得2t+t-3=0,即t2-3t+2=0,解得t=1或t=2.当t=1时,可得log2x=1,即x=2;当t=2时,可得log2x=2,即x=4.经检验x=2,x=4均符合题意.故原方程的解为x=2或x=4.14.(2021湖南长沙高一期末)某工厂新购置并安装了先进的废气处理设备,使产生的废气经过过滤后排放,以降低对空气的污染.已知过滤过程中废气的污染物数量P(单位:mg/L)与过滤时间t(单位:h)间的关系为P(t)=P0e-kt(P0,k均为非零常数,e为自然对数的底数),其中P0为t=0时的污染物数量.若经过5 h过滤后还剩余90%的污染物.(1)求常数k的值;(2)试
9、计算污染物减少到30%至少需要多长时间.(精确到1 h)(参考数据:ln 0.2-1.61,ln 0.3-1.20,ln 0.4-0.92,ln 0.5-0.69,ln 0.9-0.11)解(1)由已知得当t=0时,P=P0;当t=5时,P=90%P0.于是有90%P0=P0e-5k,解得k=-15ln 0.9(或k0.022).(2)由(1)知P=P0e(15ln0.9)t,当P=30%P0时,有0.3P0=P0e(15ln0.9)t,解得t=ln0.315ln0.9-1.2015(-0.11)=60.1155.故污染物减少到30%至少需要55 h.新情境创新练15.已知2ylogy4-2y-1=0(y0,y1),logx5xlog5x=-1(x0,x1),是否存在一个正数P,使得P=1x-y?解存在.由2ylogy4-2y-1=0,得2ylogy4-12=0,logy4=12,即y=16.由logx5xlog5x=-1,得logx5x=-1log5x,即logx5x=-logx50.12(logx5+1)=(logx5)2,整理得2(logx5)2-logx5-1=0,解得logx5=-12(logx5=1舍去),1x=25.从而P=1x-y=25-16=3,即存在一个正数P=3,使得P=1x-y成立.5
