文2-6對(duì)數(shù)與對(duì)數(shù)函數(shù)

1、對(duì)數(shù)與對(duì)數(shù)函數(shù)一、對(duì)數(shù)式一、對(duì)數(shù)式1.1.對(duì)數(shù)的定義對(duì)數(shù)的定義 如果如果ax=N(a0且且a1)，那么數(shù)，那么數(shù)x叫做叫做以以a為底為底N的對(duì)數(shù)的對(duì)數(shù),記記作作_,其中其中_叫做對(duì)數(shù)的底叫做對(duì)數(shù)的底數(shù)數(shù),_叫做真數(shù)叫做真數(shù). x x=log=loga aN Na aN N對(duì)數(shù)與指數(shù)的互化對(duì)數(shù)與指數(shù)的互化ax=Nx x=log=loga aN N推論：推論： = = _;_;logloga aa aN N =_(=_(a a00且且a a1). 1). NaalogN NN N對(duì)數(shù)形式對(duì)數(shù)形式特點(diǎn)特點(diǎn)記法記法一般對(duì)數(shù)一般對(duì)數(shù)底數(shù)為底數(shù)為a a( (a a00且且a a1)1)_常用對(duì)數(shù)常用對(duì)數(shù)底

2、數(shù)為底數(shù)為_(kāi)自然對(duì)數(shù)自然對(duì)數(shù)底數(shù)為底數(shù)為_(kāi)logloga aN N1010lg lg N Ne eln ln N N2.2.幾種常見(jiàn)對(duì)數(shù)幾種常見(jiàn)對(duì)數(shù)3.3.對(duì)數(shù)的性質(zhì)對(duì)數(shù)的性質(zhì) loga1=0(a0且且a1). logaa=1(a0且且a1) 零和負(fù)數(shù)沒(méi)有對(duì)數(shù)。零和負(fù)數(shù)沒(méi)有對(duì)數(shù)。4.4.對(duì)數(shù)的運(yùn)算法則對(duì)數(shù)的運(yùn)算法則 如果如果a a00且且a a1,1,M M0,0,N N0,0,那么那么 logloga a( (MNMN)=_;)=_; =_; =_;logloga aM M+log+loga aN Nlogloga aM M-log-loga aN NNMalog logloga aM Mn

3、 n= = _(_(n nR R);); n nlogloga aM M .loglogMmnManam換底公式換底公式: : ( (a a, ,b b均大于零且不等于均大于零且不等于1)1)； 推廣推廣logloga ab bloglogb bc cloglogc cd d=_.=_.5.5.對(duì)數(shù)的重要公式對(duì)數(shù)的重要公式bNNaablogloglog,log1logabbalogloga ad d二、對(duì)數(shù)函數(shù)二、對(duì)數(shù)函數(shù)1.1.對(duì)數(shù)函數(shù)的定義對(duì)數(shù)函數(shù)的定義 函數(shù)函數(shù) y=logax(a0, 且且a 1)叫做叫做對(duì)對(duì)數(shù)函數(shù)數(shù)函數(shù), 其中其中 x 是自變量是自變量, 函函數(shù)的定義是數(shù)的定義是 (

4、0,+)(0,+)說(shuō)明說(shuō)明：對(duì)數(shù)函數(shù)有以下特點(diǎn)：對(duì)數(shù)函數(shù)有以下特點(diǎn)：（1）自變量在真數(shù)上，且系數(shù)為）自變量在真數(shù)上，且系數(shù)為1；（2）底數(shù)是常數(shù)，且大于）底數(shù)是常數(shù)，且大于0不等于不等于1；（3）對(duì)數(shù)式前面的系數(shù)為）對(duì)數(shù)式前面的系數(shù)為1。 稱以稱以10為底的對(duì)數(shù)函數(shù)為底的對(duì)數(shù)函數(shù)y=lgx為為常用對(duì)數(shù)函數(shù)常用對(duì)數(shù)函數(shù) 稱以無(wú)理數(shù)稱以無(wú)理數(shù)e為底的對(duì)數(shù)為底的對(duì)數(shù)函數(shù)函數(shù)y=lnx為為自然對(duì)數(shù)函自然對(duì)數(shù)函數(shù)數(shù)a a1100a a111時(shí)時(shí),_,_當(dāng)當(dāng)00 x x111時(shí)時(shí),_,_當(dāng)當(dāng)00 x x100y y00y y000增函數(shù)增函數(shù)減函數(shù)減函數(shù)注意:(1)對(duì)數(shù)函數(shù)的圖象都經(jīng)過(guò)點(diǎn)(1,0)且圖象都

5、在第一、四象限；(2)對(duì)數(shù)函數(shù)都以y軸為漸近線(當(dāng)0a1時(shí)，圖象向下無(wú)限接近y軸)；(3)同真數(shù)的對(duì)數(shù)值大小關(guān)系如圖所示，對(duì)應(yīng)關(guān)系為ylogax，ylogbx，ylogcx，ylogdx，則作直線y1，得0cd1ab，即圖象在x軸上方的部分自左向右底數(shù)逐漸增大.(5)常見(jiàn)對(duì)數(shù)方程或?qū)?shù)不等式的解法：常用同底法，利用對(duì)數(shù)函數(shù)的單調(diào)性求解.考向一考向一 對(duì)數(shù)的化簡(jiǎn)與求值對(duì)數(shù)的化簡(jiǎn)與求值【例例1 1】(1)(1)化簡(jiǎn)化簡(jiǎn): : (2)(2)化簡(jiǎn)化簡(jiǎn): : (3)(3)已知已知logloga a2=2=m m,log,loga a3=3=n n, ,求求a a2 2m m+ +n n的值的值. . ;

6、40lg50lg8lg5lg2lg;24log35 . 0解解 (1)(1)原式原式= =(2)(2)(3)(3)方法一方法一 logloga a2=2=m m,a am m=2.=2.logloga a3=3=n n,a an n=3.=3.故故a a2 2m m+ +n n=(=(a am m) )2 2a an n=4=43=12.3=12.方法二方法二 logloga a2=2=m m,log,loga a3=3=n n, ,. 145lg45lg4050lg852lg. 241828282822241log4log4log4log34log322215 . 05 . 0.1212lo

7、g3log2log22aaaaaanm知能遷移知能遷移1 1 (1)(1)化簡(jiǎn)化簡(jiǎn)(log(log4 43+log3+log8 83)(log3)(log3 32+log2+log9 92)=2)= _. _. 解析解析 .45)2log23()3log65()22(log)33(log)2log212)(log3log313log21(32213312123322原式45(2)(2)已知已知3 3a a=5=5b b= =A A, ,且且 則則A A的值是（的值是（ ） A.15 B. C. D.225 A.15 B. C. D.225解析解析 3 3a a=5=5b b= =A A,a a

8、=log=log3 3A A, ,b b=log=log5 5A A, , =log =logA A3+log3+logA A5=log5=logA A15=2,15=2, A A2 2=15,=15,A A= = 或或A A= = （舍）（舍）. . , 211ba1515ba111515BCA比較下列各組數(shù)的大小比較下列各組數(shù)的大小. . (1) (1) (2)log (2)log1.11.10.70.7與與loglog1.21.20.7;0.7; (3) (3)已知已知 比較比較2 2b b,2,2a a,2,2c c的大的大 小關(guān)系小關(guān)系. .;56log32log53與,loglog

9、log212121cab考向三比較大小，解不等式考向三比較大小，解不等式解解 （1 1） loglog1=0, log5 51=0,1=0, 32log356log5.56log32log53(2)(2)方法一方法一 00.71,1.1log00.71,1.1log0.70.71.1log1.1log0.70.71.2,1.2,即由換底公式可得即由換底公式可得loglog1.11.10.7log0.7log1.21.20.7.0.7.方法二方法二 作出作出y y=log=log1.11.1x x與與y y=log=log1.21.2x x的圖象的圖象. .如圖所示兩圖象與如圖所示兩圖象與x x=0.7=0.7相相交可知交可知loglog1.11.10.7log0.7 a a c c, ,而而y y=2=2x x是增函數(shù)，是增函數(shù)，2 2b b22a a22c c. . ,2 . 1log11 . 1log17 . 07 . 0 xy21log,logloglog212121cab且比較對(duì)數(shù)式的大小，解決此類(lèi)問(wèn)題的方法很多比較對(duì)數(shù)式的大小，解決此類(lèi)問(wèn)題的方法很多當(dāng)當(dāng)?shù)讛?shù)相同底數(shù)相同時(shí)可直接利用對(duì)數(shù)函數(shù)的單