中职数学4.3.2对数的运算法则

等价关系：负数和零没有对数结论：指数式对数式(1)常用对数：log10N=lgN(2)自然对数：logeN=lnN（e=2.71828······）两个重要的对数：知识回顾baN(0,1,0)aaNlogaalog1a01(0,1,0)aaNlogaNb(,)(,)()(,)()()mnmnmnnaamnRamnRaamnRabnR指数运算法则知识回顾mnamnamnannab问题一、研究以下两组对数求值结论求值2log42log82log(48)2log(48)3log93log(927)3log(927)3log272log82log4+3log273log9+235235探究一、两个正数的积的对数，等于同一底数的这两个数的对数的和。------（积的对数）log()loglogaaaMNMN（1）（积的对数等于同底对数之和）即baNlogaNb(0,1,0)aaNlog()loglogaaaMNMN证明：pMaqNalogaMplogaNq设，，根据对数的定义得pqpqMNaaa所以，log(MN)logpqaaapq根据对数的定义得，log()loglogaaaMNMN所以，证明：log()loglogaaaMNMN2log(416)3(2)log(327)121(1)log(2)4(3)lg4lg2566(4)log12log3（积的对数）练习:例题讲解：lg5lg20lg(520)lg100222log4log162461422问题二、研究以下两组对数求值结论求值2log642log16264log16264log165log255log125525log125525log1252log162log64-5log255log125-64223-1探究二、两个正数的商的对数，等于同一底数的被除数的对数减去除数的对数。-----（商的对数）logloglogaaaMMNN即（2）（商的对数等于同底对数差）baNlogaNb(0,1,0)aaN证明：logloglogaaaMMNNpaMqaN，根据对数的定义得logaMplogaNq设，ppqqMaaNa所以，loglogpqaaMapqN根据对数的定义得，logloglogaaaMMNN所以，证明：logloglogaaaMMNN28(1)log433(2)log72log8（商的对数）练习：例题讲解：264(1)log422log64log462433(2)log5log1535log1531log3112问题三、研究以下两组数据求值结论求值3log923log412312log9_____32log42log431log92123log9323log4_____6611探究三、一个正数的幂的对数，等于幂指数乘以这个数的对数。------（幂的对数）loglogbaaMbM即（3）（幂的对数等于幂指数乘以此数的对数）baNlogaNb(0,1,0)aaN证明：loglogbaaMbMpaM证明：logaMp设，根据对数的定义得()bpbpbMaa所以，loglogbpbaaMapb根据对数的定义得loglogbaaMbM所以，loglogbaaMbM32log823(2)log275(4)lg102(3)lne351(1)log()25（幂的对数）练习：23log4例题讲解：23log833932632log4235log(5)65log(5)56log56513log()25253log5523log56625logbaalogabab对数的运算性质：log(MN)loglogaaaMNlogloglogaaaMMNNloglogbaaMbM0,10,0aaMN且对数运算性质的综合运用：269(1)log(93)55(1)log20log4131(4)lg100log8123(3)log(279)40.8(4)lg10lnlog1e2(2)lnloge1、例题讲解：2、练习：2699log9log3235242loga(M·N)＝logaM十logaNNMloga＝logaM－logaNlogaMn＝nlogaM积、商、幂的对数运算法则：alog(MN)loglog()logloglog()loglog()aaaaabaaMNMMNNMbM1230,10,0aaMN且课堂小结课堂作业：P109练习2、习题1课后作业：本节练习册预习新课：§4.6对数函数性质补充：logaNaNlogloglogcacNNa2log3(2)24log5(1)422log3(3)223(4)log9log4(0,1;0;0,1)aaNcc练习：（证明）（证明）证明：log(0,1,0)aNaNaaN证明：logaNaxlogloglogaNaaxaloglogaaxNxNlogaNaN设在等式两边取对数，即所以，即证明：logloglogcacNNa(0,1,0,0,1)aaNcc证明：，则xaNlogaNx设loglogxccaNc等式两边取以为底的对数loglogccxaN即loglogccNxa所以，logloglogcacNNa即