ab=Nb=logaN指数式对数式1.关系:2.特殊对数:1)常用对数—以10为底的对数;lgN2)自然对数—以e为底的对数;lnN4.对数恒等式:NaNalog3.重要结论:1)logaa=1;2)loga1=0aNb指数式Nab底数幂指数对数式bNalog对数的底数真数对数积、商、幂的对数运算法则:如果a0,a1,M0,N0有:)()()(3R)M(nnlogMlog2NlogMlogNMlog1NlogMlog(MN)loganaaaaaaa新课教学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2用,logxa,logyazalog表示下列各式:32log)2(;(1)logzyxzxyaa例题讲解其他重要公式1:NmnNanamloglog证明:设,logpNnam由对数的定义可以得:,)(pmnaN∴即证得NmnNanamloglogmpnaNpnmNalogpnmaN换底公式2aNNccalogloglog)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例1计算(1))42(log752(3)27log9例题讲解(4)8log7log3log732(2)5100lg(1)18lg7lg37lg214lg例2计算:解法一:18lg7lg37lg214lg18lg7lg)37lg(14lg218)37(714lg201lg)32lg(7lg37lg2)72lg(2)3lg22(lg7lg)3lg7(lg27lg2lg018lg7lg37lg214lg解法二:例题讲解(2)例3计算:9lg243lg3lg23lg525解:1023lg)10lg(32lg)3lg(2.1lg10lg38lg27lg)3(2213213253lg3lg9lg243lg)2(2.1lg10lg38lg27lg)3(12lg23lg)12lg23(lg2323例题讲解积、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课堂小结例4已知,求的值.a12log324log3312a例5设,已知,求的值.35abm112abm15例题讲解.45求lg,a2lg已知b,lg3例6:练习:已知log23=a,log37=b,用a,b表示log4256解:因为log23=a,则,又∵log37=b,∴2log13a1312log7log2log37log42log56log56log33333342babab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课堂小结2、计算:①②3log12.054219432log2log3log1、计算:(1)log535-2log5+log57-log51.837(2)lg25+lg2lg5+lg2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