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对数的运算性质一般地,如果1,0aa的b次幂等于N,就是Nab,那么数b叫做以a为底N的对数,记作bNaloga叫做对数的底数,N叫做真数。定义:复习上节内容a例如:1642216log41001022100log102421212log401.0102201.0log10复习上节内容有关性质:⑴负数与零没有对数(∵在指数式中N0)⑵,01loga1logaa⑶对数恒等式NaNalog)1(复习上节内容),1,0(log)2(Rbaababa⑷常用对数:我们通常将以10为底的对数叫做常用对数。为了简便,N的常用对数N10log简记作lgN。⑸自然对数:在科学技术中常常使用以无理数e=2.71828……为底的对数,以e为底的对数叫自然对数。为了简便,N的自然对数Nelog简记作lnN。(6)底数a的取值范围:),1()1,0(真数N的取值范围:),0(复习上节内容一创设情境:•指数幂运算有那些性质?)()(),()(),(RnbaabRnmaaRnmaaannnmnnmnmnm对数运算也有相应的运算性质吗?如果有,它们之间有什么样的联系呢?)()(),()(),(RnbaabRnmaaRnmaaannnmnnmnmnm三师生探究:积、商、幂的对数运算性质:如果a0,a1,M0,N0有:)()()(3R)M(nnlogMlog2NlogMlogNMlog1NlogMlog(MN)loganaaaaaaa为了证明以上公式,请同学们再回顾一下指数运算性质:证明:①设,logpMa,logqNa由对数的定义可以得:,paMqaN∴MN=paqaqpaqpMNalog即证得)(1NlogMlog(MN)logaaa证明:②设,logpMa,logqNa由对数的定义可以得:,paMqaN∴qpaaqpaqpNMalog即证得NM)(2NlogMlogNMlogaaa证明:③设,logpMa由对数的定义可以得:,paM∴npnaMnpMnalog即证得)(3R)M(nnlogMlogana上述证明是运用转化的思想,先通过假设,将对数式化成指数式,并利用幂的运算性质进行恒等变形;然后再根据对数定义将指数式化成对数式。)()()(3R)M(nnlogMlog2NlogMlogNMlog1NlogMlog(MN)loganaaaaaaa①简易语言表达:“积的对数=对数的和”……②有时逆向运用公式③真数的取值范围必须是),0(④对公式容易错误记忆,要特别注意:,loglog)(logNMMNaaaNMNMaaaloglog)(log例1讲解范例解(1)解(2)用,logxa,logyazalog表示下列各式:32log)2(;(1)logzyxzxyaazxyzxyaaalog)(loglog3121232log)(loglogzyxzyxaaazyxaaalogloglog31212logloglogzyxaaazyxaaalog31log21log21.用lgx,lgy,lgz表示下列各式:练习(1)(4)(3)(2))lg(xyzzxy2lgzxy3lg=lgx+2lgy-lgz;zyx2lg=lgx+lgy+lgz;=lgx+3lgy-21lgz;zyxlglg2lg21例2计算(1)(2))42(log75227log3讲解范例解:)42(log752522log724log522log1422log=5+14=19解:27log3333log3log333练习(1)(4)(3)(2)2.求下列各式的值:15log5log332lg5lg31log3log553log6log2236log2)25lg()313(log5155log32log2110lg11log50133log1巩固练习)24(log)6(57227log3log)1(995100lg)2(5lg241lg)3()44(log)4(2100lg100000lg)5(1.计算22/5-235/219的式子表示2.已知用,ba5log,3log22ba,6.0log)1(230log)2(21253log)3(42a-b½(1+a+b)a/(12b)巩固练习小结:积、商、幂的对数运算法则:如果a0,a1,M0,N0有:)()()(3R)M(nnlogMlog2NlogMlogNMlog1NlogMlog(MN)loganaaaaaaa其他重要公式:NmnNanamloglogaNNccalogloglog)0),,1()1,0(,(Nca1loglogabba),1()1,0(,ba其他重要公式2:aNNccalogloglog)0),,1()1,0(,(Nca证明:设由对数的定义可以得:,paN即证得pNalog,loglogpccaN,loglogapNccaNpccloglogaNNccalogloglog这个公式叫做换底公式讲解范例(3)8log7log3log732解:8log7log3log7322lg3lg2lg2lg32lg2lg3=33lg7lg7lg8lg(1)18lg7lg37lg214lg例3计算:讲解范例解法一:18lg7lg37lg214lg18lg7lg)37lg(14lg218)37(714lg201lg)32lg(7lg37lg2)72lg(2)3lg22(lg7lg)3lg7(lg27lg2lg018lg7lg37lg214lg解法二:(2)例3计算:讲解范例9lg243lg3lg23lg525解:1023lg)10lg(32lg)3lg(2.1lg10lg38lg27lg)3(2213213253lg3lg9lg243lg)2(2.1lg10lg38lg27lg)3(12lg23lg)12lg23(lg2323其他重要公式3:abbalog1log),1()1,0(,ba证明:由换底公式取以b为底的对数得:还可以变形,得,1logbbaNNccalogloglogabbbbalogloglogabbalog1log1loglogabba其他重要公式1:NmnNanamloglog证明:设,logpNnam由对数的定义可以得:,)(pmnaN∴即证得NmnNanamloglogmpnaNpnmNalogpnmaN再见
本文标题:积商幂的对数
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