新教材高中數(shù)學(xué)第4章冪函數(shù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)3.1對(duì)數(shù)的概念3.2對(duì)數(shù)的運(yùn)算法則課件湘教版必修第一冊(cè)

4.3

對(duì)數(shù)函數(shù)4.3.1

對(duì)數(shù)的概念1|對(duì)數(shù)的概念4.3.2

對(duì)數(shù)的運(yùn)算法則1.對(duì)數(shù)的概念如果ab=N(a>0且a≠1),那么b叫作以a為底,(正)數(shù)N的對(duì)數(shù),記作①

b=logaN

.這

里a叫作對(duì)數(shù)的②底數(shù)

,N叫作對(duì)數(shù)的真數(shù).2.常用對(duì)數(shù)與自然對(duì)數(shù)通常,我們將以10為底的對(duì)數(shù)叫作對(duì)數(shù)的常用對(duì)數(shù),并把log10N記為③

lgN

;以e

(e=2.71828…)為底的對(duì)數(shù)稱為自然對(duì)數(shù),并把logeN記為④

lnN

.3.對(duì)數(shù)的基本恒等式(1)

=N(N>0,a>0且a≠1);(2)b=logaab(b∈R,a>0且a≠1).特別地,logaa=logaa1=1,loga1=logaa0=0.如果a>0,且a≠1,M>0,N>0,那么(1)loga(MN)=⑤

logaM+logaN

;(2)logaMn=⑥

nlogaM

(n∈R);(3)loga

=logaM-logaN.2|對(duì)數(shù)的運(yùn)算法則1.公式:logbN=

.2.推論:logab=

,

bm=

logab(a>0,且a≠1;b>0,且b≠1;n≠0).3|換底公式1.因?yàn)?-2)2=4,所以2=log(-2)4.

(

?)提示:因?yàn)閷?duì)數(shù)的底數(shù)a應(yīng)滿足a>0且a≠1,所以結(jié)論錯(cuò)誤.2.因?yàn)閍1=a(a>0且a≠1),所以logaa=1.

(√)3.若lnN=

,則N=

.

(

?)提示:lnN=

,則N=

.4.logab·logbc·logcd=logad(a>0,b>0,c>0,d>0,且a≠1,b≠1,c≠1).

(√)提示:根據(jù)對(duì)數(shù)換底公式知結(jié)論正確.5.使對(duì)數(shù)log4(2a+1)有意義的a的取值范圍是

.

(√)提示:要使對(duì)數(shù)log4(2a+1)有意義,需使2a+1>0,解得a>-

,故結(jié)論正確.判斷正誤，正確的畫“√”，錯(cuò)誤的畫“?”.1|利用對(duì)數(shù)的運(yùn)算性質(zhì)化簡(jiǎn)、求值1.利用對(duì)數(shù)的運(yùn)算性質(zhì)求值的解題關(guān)鍵是化異為同,先使各項(xiàng)底數(shù)相同,再找真數(shù)

間的聯(lián)系.2.對(duì)于復(fù)雜的運(yùn)算式,可先化簡(jiǎn)再計(jì)算.化簡(jiǎn)的常用方法:①“拆”:將積(商)的對(duì)

數(shù)拆成兩對(duì)數(shù)之和(差);②“收”:將同底對(duì)數(shù)的和(差)收成積(商)的對(duì)數(shù).3.在利用換底公式進(jìn)行化簡(jiǎn)求值時(shí),一般情況下是根據(jù)題中所給對(duì)數(shù)式的具體特

點(diǎn)選擇恰當(dāng)?shù)牡讛?shù)進(jìn)行換底,如果所給的對(duì)數(shù)式中的底數(shù)和真數(shù)互不相同,我們

可以選擇以10為底數(shù)進(jìn)行換底.利用換底公式化簡(jiǎn)與求值的思路:①用對(duì)數(shù)的運(yùn)算性質(zhì)進(jìn)行部分運(yùn)算

換成同一底數(shù).②統(tǒng)一換為常用對(duì)數(shù)(或自然對(duì)數(shù))

化簡(jiǎn)、求值.(1)化簡(jiǎn):

;(2)已知log189=a,18b=5,用a,b表示log3645.解析

(1)原式=

·

=lo

·lo

9=lo

·3lo

32=-

·log32·3log23=-

.(2)解法一:∵18b=5,∴b=log185,于是log3645=

=

=

=

=

.解法二:∵18b=5,∴b=log185,于是log3645=

=

=

.解法三:∵log189=a,18b=5,∴l(xiāng)g9=alg18,lg5=blg18,∴l(xiāng)og3645=

=

=

=

=

.2|對(duì)數(shù)運(yùn)算性質(zhì)的綜合應(yīng)用1.在對(duì)數(shù)式、指數(shù)式的互化運(yùn)算中,要注意靈活運(yùn)用定義和運(yùn)算性質(zhì),尤其要注意

條件和結(jié)論之間的關(guān)系.2.解決對(duì)數(shù)應(yīng)用問題時(shí),首先要理解題意,弄清關(guān)鍵詞及字母的含義,然后恰當(dāng)設(shè)

未知數(shù),建立數(shù)學(xué)模型,最后轉(zhuǎn)化為對(duì)數(shù)問題求解.已知a,b,c是不等于1的正數(shù),且ax=by=cz,

+

+

=0,求abc的值.思路點(diǎn)撥設(shè)ax=by=cz=t,則x=logat,y=logbt,z=logct,代入

+

+

=0并用對(duì)數(shù)的運(yùn)算性質(zhì)可求得abc的值,也可以用換底公式進(jìn)行計(jì)算.解析

解法一:設(shè)ax=by=cz=t,則t>0,且t≠1,則x=logat,y=logbt,z=logct,∴

+

+

=

+

+

=logta+logtb+logtc=logt(abc)=0,∴abc=t0=1,即abc=1.即abc=1.解法二:設(shè)ax=by=cz=t,則t>0,且t