2022年对数函数知识点总结 .pdf
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1、学习必备精品知识点对数函数(一)对数1对数的概念:一般地,如果Nax)1,0(aa,那么数x叫做以a为底N的对数,记作:Nxalog(a 底数,N 真数,Nalog 对数式)说明:1注意底数的限制0a,且1a;2xNNaaxlog;3注意对数的书写格式两个重要对数:1常用对数:以10 为底的对数Nlg;2自然对数:以无理数71828.2e为底的对数的对数Nln(二)对数的运算性质如果0a,且1a,0M,0N,那么:1Ma(log)NMalogNalog;2NMalogMalogNalog;3naMlognMalog)(Rn注意:换底公式abbccalogloglog(0a,且1a;0c,且1c
2、;0b)利用换底公式推导下面的结论(1)bmnbanamloglog;(2)abbalog1log(二)对数函数1、对数函数的概念:函数0(logaxya,且)1a叫做对数函数,其中x是自变量,函数的定义域是(0,+)注意:1对数函数的定义与指数函数类似,都是形式定义,注意辨别。如:xy2log2,5log5xy都不是对数函数,而只能称其为对数型函数2对数函数对底数的限制:0(a,且)1a2、对数函数的性质:a1 0a0 得0 x,函数2logxya的定义域是0 x x;(2)由04x得4x,函数)4(logxya的定义域是4x x;(3)由9-02x得-33x,函 数)9(l o g2xya
3、的 定 义 域 是33xx例 2求函数251xy和函数22112xy)0(x的反函数。解:(1)125xy115()log(2)fxx(-2)x;(2)211-22xy-112()log(-2)fxx5(2)2x例 4比较下列各组数中两个值的大小:(1)2log 3.4,2log 8.5;(2)0.3log1.8,0.3log2.7;(3)log 5.1a,log 5.9a.解:(1)对数函数2logyx在(0,)上是增函数,于是2log 3.42log 8.5;(2)对数函数0.3logyx在(0,)上是减函数,于是0.3log1.80.3log2.7;(3)当1a时,对数函数logayx在
4、(0,)上是增函数,于是log 5.1alog 5.9a,当1oa时,对数函数logayx在(0,)上是减函数,于是log 5.1alog 5.9a例 5比较下列比较下列各组数中两个值的大小:(1)6log 7,7log 6;(2)3log,2log 0.8;(3)0.91.1,1.1log0.9,0.7log0.8;(4)5log 3,6log 3,7log 3文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF
5、1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S
6、6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编
7、码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10
8、P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z
9、7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU
10、2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K
11、1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1学习必备精品知识点解:(1)66log 7log 61,77log 6log 71,6log 77log 6;(2)33loglog 10,22log 0.8log 10,3log2log 0.8 (3)0.91.11.11,1.11.1log0.9log10,0.70.70.70log1log0.8log0.71,0.91.10.7log0.81.1log0.9(4)3330log 5log 6l
12、og 7,5log 36log 37log 3例 7求下列函数的值域:(1)2log(3)yx;(2)22log(3)yx;(3)2log(47)ayxx(0a且1a)解:(1)令3tx,则2logyt,0t,yR,即函数值域为R(2)令23tx,则03t,2log 3y,即函数值域为2(,log 3(3)令2247(2)33txxx,当1a时,log3ay,即值域为log3,)a,当01a时,log 3ay,即值域为(,log3a例 8判断函数22()log(1)f xxx的奇偶性。解:21xx恒成立,故()f x的定义域为(,),22()log(1)fxxx221log1xx222221l
13、og(1)xxxx22log1()xxf x,所以,()f x文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V
14、6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H
15、1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K
16、8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R
17、9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P
18、1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:
19、CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1学习必备精品知识点为奇函数。例 9求函数2132log(32)yxx的单调区间。解:令223132()24
20、uxxx在3,)2上递增,在3(,2上递减,又2320 xx,2x或1x,故232uxx在(2,)上递增,在(,1)上递减,又132logyu为减函数,所以,函数2132log(32)yxx在(2,)上递增,在(,1)上递减。例 10若函数22log()yxaxa在区间(,13)上是增函数,a的取值范围。解:令2()ug xxaxa,函数2logyu为减函数,2()ug xxaxa在 区 间(,13)上 递 减,且 满 足0u,132(13)0ag,解得22 32a,所以,a的取值范围为22 3,2【例1】(1)y=log(2)y=11log(a0a1)(3)f(x)01y=flog(3x)1
21、2a13求函数的定义域求函数,且 的定义域已知函数的定义域是,求函数的定义3221xxxa()文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文
22、档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ
23、10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B
24、9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6
25、HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N1K1K8 ZF1R9T4S6P1文档编码:CJ10P10B9Z7V6 HU2H1N
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