223對(duì)數(shù)的換底公式

1、 1、能較熟練地運(yùn)用對(duì)數(shù)運(yùn)算法則、能較熟練地運(yùn)用對(duì)數(shù)運(yùn)算法則解決問(wèn)題解決問(wèn)題; 2、加強(qiáng)數(shù)學(xué)應(yīng)用意識(shí)的訓(xùn)練，、加強(qiáng)數(shù)學(xué)應(yīng)用意識(shí)的訓(xùn)練， 提高解決應(yīng)用問(wèn)題的能力。提高解決應(yīng)用問(wèn)題的能力。積、商、冪的對(duì)數(shù)運(yùn)算法則：積、商、冪的對(duì)數(shù)運(yùn)算法則：如果如果a0，且，且a1，M0，N0有：有：NMMNaaaloglog)(logNMNMaaalogloglogR)M(nnManaloglogMnPMManPanpalogloglog)(logRnnananaaaaMMMMMlogloglog)M(log21n21MMaalog1log一、對(duì)數(shù)的換底公式一、對(duì)數(shù)的換底公式: 如何證明呢如何證明呢?aNNcca

2、logloglog)0), 1()1 , 0(,( Nca證明證明：設(shè)：設(shè) 由對(duì)數(shù)的定義可以得：由對(duì)數(shù)的定義可以得： paN 即證得即證得 pNalogpccaNloglogapNccloglogaNpccloglogaNNccalogloglog通過(guò)換底公式，人們通過(guò)換底公式，人們可以把其他底的對(duì)數(shù)可以把其他底的對(duì)數(shù)轉(zhuǎn)換為以轉(zhuǎn)換為以10或或e為底為底的對(duì)數(shù)，經(jīng)過(guò)查表就的對(duì)數(shù)，經(jīng)過(guò)查表就能求出任意不為能求出任意不為1的的正數(shù)為底的對(duì)數(shù)。正數(shù)為底的對(duì)數(shù)。二、幾個(gè)重要的推論二、幾個(gè)重要的推論: 如何證明呢如何證明呢?abbalog1logNmnNanamloglog), 1 () 1 , 0(,b

3、a證明證明：利用換底公式得：利用換底公式得：即證得即證得 NmnNanamlogloglglglgloglglglgmnaNnNnNnNamamamlogaNmnaNlglg證明證明:由換底公式由換底公式 abbalog1log即即 abbaloglog1lglglglgbaab1logloglogacbcba推論推論:例例1:計(jì)算計(jì)算:解解: 27log19 27log19333log23log23323 8log7log3log2732 9lg212log1100333 9lg212log1100333 8log7log3log27322lg2lg32lg3lg3lg7lg7lg8lg3解

4、解:例例1:計(jì)算計(jì)算: 27log19 8log7log3log2732解解: 9lg212log11003339lg2122log103339lg102392315 9lg212log1100333例例1:計(jì)算計(jì)算: 27log19 8log7log3log2732解解:.)21(2,10054:2的值求設(shè)例baba10054ba10log10log100log22242a2log224log245log100log55255b2log1110log12)21(252ba25log2log22log5log12log210105510. 9log,7log,5log:33539表示試用已知例nmnm解解:7log, 5log215log5log33392nm7log,25log33nmnm227log5log235log23log29log3333535.,07lg5lglg)7lg5(lglg:421212xxxxxx求的兩根分別為方程例07lg5lglg)7lg5(lglg2xx解解:7lg5lglglg)7lg5(lglglg2121xxxx351lg