20111014高一数学(2.2.3对数的换底公式)

积、商、幂的对数运算法则：如果a＞0，且a≠1，M＞0，N＞0有：NMMNaaaloglog)(logNMNMaaalogloglogR)M(nnManaloglogMnPMManPanpalogloglog)(logRnnananaaaaMMMMMlogloglog)M(log21n21MMaalog1log一、对数的换底公式:如何证明呢?aNNccalogloglog)0),,1()1,0(,(Nca证明：设由对数的定义可以得：paN即证得pNalogpccaNloglogapNccloglogaNpccloglogaNNccalogloglog通过换底公式，人们可以把其他底的对数转换为以10或e为底的对数，经过查表就能求出任意不为1的正数为底的对数。二、几个重要的推论:如何证明呢?abbalog1logNmnNanamloglog),1()1,0(,ba证明：利用换底公式得：即证得NmnNanamlogloglglglgloglglglgmnaNnNnNnNamamamlogaNmnaNlglg证明:由换底公式abbalog1log即abbaloglog1lglglglgbaab1logloglogacbcba推论:例1:计算:解:27log1927log19333log23log233238log7log3log27329lg212log11003339lg212log11003338log7log3log27322lg2lg32lg3lg3lg7lg7lg8lg3解:例1:计算:27log198log7log3log2732解:9lg212log11003339lg2122log103339lg1023923159lg212log1100333例1:计算:27log198log7log3log2732解:.)21(2,10054:2的值求设例baba10054ba10log10log100log22242a2log224log245log100log55255b2log1110log12)21(252ba25log2log22log5log12log210105510.9log,,7log,5log:33539表示试用已知例nmnm解:7log,5log215log5log33392nm7log,25log33nmnm227log5log235log23log29log3333535.,,07lg5lglg)7lg5(lglg:421212xxxxxx求的两根分别为方程例07lg5lglg)7lg5(lglg2xx解:7lg5lglglg)7lg5(lglglg2121xxxx351lg35lg35lglg121xx35121xx作业：1、2、