对数运算法则

对数运算法则对数的概念x(0,1)xaNaa且logaxNa一般地，若，那么数叫做以a为底N的对数，记作叫做对数的底数，N叫做真数.ab＝NlogaN＝b复习logxaaNNx底数指数真数底数对数幂对数的性质：⑴负数与零没有对数⑵⑶⑷,01loga1logaaNaNalog⑸xaxalog10loglgNN记为；eloglnNN记为；常用对数自然对数说说对数和指数间的关系吗？axN0,1logxaaaaNxN当时，xaN指数logaxN对数底数指数幂底数真数对数333log1log3log27lnlg1007lg142lglg7lg183e43?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证明:logloglogaaaMNMN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logloglogaaaMMNNloglog()naaMnMnR语言表达:一个正数的n次方的对数等于这个正数的对数n倍如果a0，a1，M0，N0有：例1计算（1）（2）)42(log7525lg100解:)42(log752522log724log522log1422log=5+14=19解:21lg1052lg105255lg100例2解（1）解（2）用,logxa,logyazalog表示下列各式：32log)2(;(1)logzyxzxyaazxyzxyaaalog)(loglog23logaxyzzyxaaalogloglog31212logloglogzyxaaazyxaaalog31log21log211232log()logaaxyz（1）18lg7lg37lg214lg例3计算：解法一：18lg7lg37lg214lg18lg7lg)37lg(14lg218)37(714lg201lg)32lg(7lg37lg2)72lg(2)3lg22(lg7lg)3lg7(lg27lg2lg018lg7lg37lg214lg解法二：例8log2322log212log21326解:原方程可化为444log(31)log(1)log(3).xxx2.解方程31(1)(3)xxx220xx21xx解得或2x方程的解是检验:1x使真数3x-1和x-1分别小于或等于0舍去1x有时可逆向运用公式真数的取值必须是(0,＋∞)注意log()aMNloglogaaMNlog()aMNloglogaaMN≠≠简易语言表达:积的对数=对数的和2、应用举例：例1、用表示下列各式：zayaxalog,log,logzxyalog)(1322zyxalog)(zayaxazaxyazxyaloglogloglog)(loglog)(1解：32322zayxazyxalogloglog)(32zayaxalogloglogxayaloglog3121xa2log例2：求下列各式的值：5100lg)2()(log52742（1）522742527421loglog(log））解：（19514225427loglog5100lg)2(5221051511001005lg)lg(lg271364232552logloglog练习：2533271315223)log(logloglog)(2325312533301)(log解：原式25502224lglglg))(lg(25105222lg)(lglg)(lg解：原式5215222lg)(lglg)(lg）（）（212221222210225222lglglglg)(lglglglglg)(lg2二换底公式abbccalogloglog证明pbalog设bappabccccloglogloglogbacpcloglogbapbaxbxa则设,logbacxcloglog两边取对数，baxccloglogabxccloglogabbccalogloglog方法二练习nabnlogbalognabmlogbmnalog1lglglglgloglogbaababba例求下列各式的值。32log9log13853222log3log32log53log32322log3log31032310adcbdcbaloglogloglog2dacdbcablglglglglglglglg1NaMaNMaloglog)(log①NaMaNMalogloglog②)(loglogRnMannMa③）公式知识回顾：（1NaNalog