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《《对数及对数函数》练习题及讲解.pdf》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、'.《对数及对数函数》练习题讲解知识梳理:b1、对数的定义:如果a(a>0,a≠1)的b次幂等于N,就是a=N,那么数b叫做a为底N的对数,记作logaN=b,a叫做对数的底数,N叫做真数。(N>0)b2、指数和对数的关系:aNlogaNblogN3、对数恒等式:∴log10aa,logaa1,aNloga(MN)logaMlogaNM4、运算法则:logalogaMlogaNNnlogaMnlogaM(nR)logca5、换底公式:logablogcb6、两个较为常用的推论:nn1logablog
2、ba12logamblogab(a,b>0且均不为1)mx7、对数函数定义:函数ylogax(a0且a1)叫做对数函数;它是指数函数ya(a0且a1)的反函数。8、对数函数图象和性质:aa>103、)9;(2)log(x1)(x8x7)1;(3)log232331(解析(1)log43,得x4或x3.(2)由对数性质得x8.x41x1(3)令xlog23=log23,∴2323,∴x1)2323lg8lg125lg2lg52例2:计算(1)计算:log155log1545+(log153)(2)lg10lg0.1222(3)lg5lg8lg51g20(lg2)32解:(1)解一:原式=log155(log153+1)+(log153)=log155+log153(log155+log153)=4、log155+log153log1515=log155+log153=log15151522解二:原式=log15log15(153)(log153)=(1-log153)(1+log153)+(log153)322=1-(log153)+(log153)=18125lg()(2)=252222lg(25)2lg104122lg102(3)原式2lg52lg2(1lg2)(1lg2)(lg2)222(lg5lg2)1(lg2)(lg2)3321变式:计算:(1)lg5lg8000(lg2)lglg05、.06(=1)64(2)(log43log83)(log32log92)log1322;.'.54解:原式(log223log233)(log32log322)log1221115535555(log23log23)(log32log32)log23log322324624442b例3:已知log9a,185,求log45.183618b解:由log189a可知alog181log182,又由185,可得2log1845log189log185abblog185,故log3645log18361lo6、g1822a变式:若log83=p,log35=q,求lg5解:∵log83=p∴log233plg33plg23p(1lg5)lg5又∵log35q∴lg5qlg33pq(1lg5)lg33pq∴(13pq)lg53pq∴lg513pq例4:比较下列各组数的大小:5.10.9(1)ln0.99与ln0.9(2)p0.9,m5.1,nlog0.95.122(3)若1xd,alogdx,blogdx,clogd(logdx).解:(1)由ylnx在0,上单调递增,且00.990.9,故ln0.997、n0.9.5.100.90(2)log0.95.1log0.910,而0.90.91,5.15.11,npm(3)令logdxu,由1xd可知0logdx1即u0,1.2则au,b2u,clogdu,u0,1,在同一坐标系下画出这三个函数的图象,如图示:可知b最大,c最小,即cab.变式:比较下列各数大小:;.'.112(1)log0.30.7与log0.40.3(2)log0.60.8,log3.40.7和3(3)log0.1和log0.10.30.2解:(1)∵log0.30.7log0.30.8、31log0.40.3log0.40.41∴log0.7log0.30.30.4112(2)∵0log0.60.81log3.40.7013112∴log3.40.7log0.60.8311(3)解:log0.30.10log0.20.10log0.10.3log0.10.2∵log0.10.3log0.10.2∴log0.30.1log0.20.1例5:求下列函数的定义域、值域:21x122(1)y2(2)ylog2(x2x5)(3)ylog1(x4x5)432(4
3、)9;(2)log(x1)(x8x7)1;(3)log232331(解析(1)log43,得x4或x3.(2)由对数性质得x8.x41x1(3)令xlog23=log23,∴2323,∴x1)2323lg8lg125lg2lg52例2:计算(1)计算:log155log1545+(log153)(2)lg10lg0.1222(3)lg5lg8lg51g20(lg2)32解:(1)解一:原式=log155(log153+1)+(log153)=log155+log153(log155+log153)=
4、log155+log153log1515=log155+log153=log15151522解二:原式=log15log15(153)(log153)=(1-log153)(1+log153)+(log153)322=1-(log153)+(log153)=18125lg()(2)=252222lg(25)2lg104122lg102(3)原式2lg52lg2(1lg2)(1lg2)(lg2)222(lg5lg2)1(lg2)(lg2)3321变式:计算:(1)lg5lg8000(lg2)lglg0
5、.06(=1)64(2)(log43log83)(log32log92)log1322;.'.54解:原式(log223log233)(log32log322)log1221115535555(log23log23)(log32log32)log23log322324624442b例3:已知log9a,185,求log45.183618b解:由log189a可知alog181log182,又由185,可得2log1845log189log185abblog185,故log3645log18361lo
6、g1822a变式:若log83=p,log35=q,求lg5解:∵log83=p∴log233plg33plg23p(1lg5)lg5又∵log35q∴lg5qlg33pq(1lg5)lg33pq∴(13pq)lg53pq∴lg513pq例4:比较下列各组数的大小:5.10.9(1)ln0.99与ln0.9(2)p0.9,m5.1,nlog0.95.122(3)若1xd,alogdx,blogdx,clogd(logdx).解:(1)由ylnx在0,上单调递增,且00.990.9,故ln0.997、n0.9.5.100.90(2)log0.95.1log0.910,而0.90.91,5.15.11,npm(3)令logdxu,由1xd可知0logdx1即u0,1.2则au,b2u,clogdu,u0,1,在同一坐标系下画出这三个函数的图象,如图示:可知b最大,c最小,即cab.变式:比较下列各数大小:;.'.112(1)log0.30.7与log0.40.3(2)log0.60.8,log3.40.7和3(3)log0.1和log0.10.30.2解:(1)∵log0.30.7log0.30.8、31log0.40.3log0.40.41∴log0.7log0.30.30.4112(2)∵0log0.60.81log3.40.7013112∴log3.40.7log0.60.8311(3)解:log0.30.10log0.20.10log0.10.3log0.10.2∵log0.10.3log0.10.2∴log0.30.1log0.20.1例5:求下列函数的定义域、值域:21x122(1)y2(2)ylog2(x2x5)(3)ylog1(x4x5)432(4
7、n0.9.5.100.90(2)log0.95.1log0.910,而0.90.91,5.15.11,npm(3)令logdxu,由1xd可知0logdx1即u0,1.2则au,b2u,clogdu,u0,1,在同一坐标系下画出这三个函数的图象,如图示:可知b最大,c最小,即cab.变式:比较下列各数大小:;.'.112(1)log0.30.7与log0.40.3(2)log0.60.8,log3.40.7和3(3)log0.1和log0.10.30.2解:(1)∵log0.30.7log0.30.
8、31log0.40.3log0.40.41∴log0.7log0.30.30.4112(2)∵0log0.60.81log3.40.7013112∴log3.40.7log0.60.8311(3)解:log0.30.10log0.20.10log0.10.3log0.10.2∵log0.10.3log0.10.2∴log0.30.1log0.20.1例5:求下列函数的定义域、值域:21x122(1)y2(2)ylog2(x2x5)(3)ylog1(x4x5)432(4
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