高一数学《对数与对数运算3》PPT课件[1]

2.2.1对数与对数运算3探究换底公式：)0;1,0;1,0(logloglogbccaababacc且且如何推导？)0;1,0;1,0(logloglogbccaababacc且且证明：.logloglog,loglog,loglog,loglog,,logabbabpbapbacbabpccacccccpcpa即所以则有为底的对数两边取以，则令其他重要公式2:NmnNanamloglog证明：利用换底公式得：即证得NmnNanamlogloglglglgloglglglgmnaNnNnNnNamamamlogaN特别地：当m=1时，naalogMnlogM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,1logbbaNNccalogloglogabbbbalogloglogabbalog1log1loglogabba归纳aNNccalogloglog1、换底公式：)0),,1()1,0(,(NcaNmnNanamloglog2、),1()1,0(,,loglog1loglog1log31babbbabnanaabann、),1()1,0(,log4anana、例1计算（1）（2）)42(log75227log9解:)42(log752522log724log522log1422log=5+14=19解:27log9333log23log23323例题讲解（3）8log7log3log732解:8log7log3log7322lg3lg2lg2lg32lg2lg3=33lg7lg7lg8lg例题讲解P684P7511