高中数学必修一教学课件：221对数运算VIP免费

对数的运算性质课前练习:333log1log3log27lnlg1007lg142lglg7lg183e⑴给出四个等式:1)lg(lg10)0;2)lg(ln)0;3)e2若lgx=10,则x=10;4)若lnx=e,则x=e其中正确的是________⑵⑶⑷?证明：①设logloglogaaaMNMN对数的运算性质证明:证明：①设,logpMa,logqNa由对数的定义可以得：,paMqaN∴MN=paqaqpaqpMNalog即证得logloglogaaaMNMN对数的运算性质证明:logloglogaaaMNMN对数的运算性质两个正数的积的对数等于这两个正数的对数和两个正数的商的对数等于这两个正数的对数差logloglogaaaMNMN⑴logloglogaaaMMNNloglog()naaMnMnR语言表达:zxxk一个正数的n次方的对数等于这个正数的对数n倍如果a>0，a1，M>0，N>0有：证明：②设,logpMa,logqNa由对数的定义可以得：,paMqaN∴qpaaqpaqpNMalog即证得NMlogloglogaaaMMNN证明:aaaMloglogMlogNN证明：设,logpMa由对数的定义可以得：,paM∴npnaMnpMnalog即证得naalogMnlogM(nR)loglognaaMnM证明:例1讲解范例解（1）解（2）zx,xk用,logxa,logyazalog表示下列各式：32log)2(;(1)logzyxzxyaa例1讲解范例解（1）解（2）用,logxa,logyazalog表示下列各式：32log)2(;(1)logzyxzxyaazxyzxyaaalog)(loglog23logaxyzzyxaaalogloglog31212logloglogzyxaaazyxaaalog31log21log211232log()logaaxyz例2计算（1）（2）)42(log7525lg100讲解范例解:解:例2计算（1）（2）)42(log7525lg100讲解范例解:)42(log752522log724log522log1422log=5+14=19zx,xk,解:21lg1052lg105255lg100对数的运算性质说明:2)有时可逆向运用公式3)真数的取值必须是(0,＋∞)4)注意log()aMNloglogaaMNlog()aMNloglogaaMN≠≠logloglogaaaMNMN⑴logloglogaaaMMNN⑵loglog()naaMnMnR⑶如果a>0，a1，M>0，N>0有：1)简易语言表达:”积的对数=对数的和”……课堂小结:1⑴若lglg2lg3lg,xabc则______x661log12log22⑵的值为______⑶22log843log843_____________巩固练习:751.2.3p提高练习:探究：NmnNanamloglogaNNccalogloglog)0),,1()1,0(,(Nca1loglogabba),1()1,0(,baloglogloglog1loglogmmnaaxnmnamxnnanaanNNmNxaNaNNmxnxNNmm证明：loglogloglog1loglogmmnaaxnmnamxnnanaanNNmNxaNaNNmxnxNNmm证明：loglogloglog,log,log,,loglogloglogqcacccaPqkpcaccbbabpaqbkbcacbabpbcqa换底公式的证明证明：loglog1loglogloglogloglogloglog1abcaccbcabbabbaaabba),1()1,0(,ba证明：