高中數(shù)學(xué)第二章基本初等函數(shù)（Ⅰ）2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算課件新人教A版必修

對(duì)數(shù)的概念一般地,如果ax=N(a>0,且a≠1),那么數(shù)x叫做以①

a

為底N的對(duì)數(shù),記作②

x=logaN

,其中a叫做對(duì)數(shù)的③底數(shù)

,N叫做④真數(shù)

.(1)負(fù)數(shù)和零沒(méi)有對(duì)數(shù).(2)特殊值:1的對(duì)數(shù)是0,即loga1=0(a>0,且a≠1);底數(shù)的對(duì)數(shù)是1,即logaa=1(a>0,且a≠1).2.2對(duì)數(shù)函數(shù)2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算(1)當(dāng)a>0,且a≠1時(shí),ax=N?⑧

x=logaN

.(2)對(duì)數(shù)恒等式:

=N;logaaN=N(a>0,且a≠1).名稱(chēng)定義記法常用對(duì)數(shù)以⑤10

為底的對(duì)數(shù)叫做常用對(duì)數(shù)⑥

lgN

自然對(duì)數(shù)以無(wú)理數(shù)e=2.71828…為底的對(duì)數(shù)稱(chēng)為自然對(duì)數(shù)⑦

lnN

對(duì)數(shù)的運(yùn)算性質(zhì)如果a>0,且a≠1,M>0,N>0,那么:(1)loga(M·N)=⑨

logaM+logaN

;(2)loga

=⑩

logaM-logaN

;(3)logaMn=

nlogaM

(n∈R).

對(duì)數(shù)換底公式及其推論(1)對(duì)數(shù)換底公式:logab=

(a>0,且a≠1;b>0;c>0,且c≠1);(2)推論:lo

bn=

logab,logab=

(a>0,且a≠1;b>0,且b≠1;m≠0).

判斷正誤,正確的畫(huà)“√”,錯(cuò)誤的畫(huà)“?”.1.以e為底的對(duì)數(shù)叫做自然對(duì)數(shù).

(√)a(xy)=logax+logay(a>0,且a≠1).

(

?)當(dāng)x<0,y<0時(shí),可知結(jié)論錯(cuò)誤.a(-2)2=2loga(-2)(a>0,且a≠1).

(

?)公式logaMn=nlogaM(n∈R)中的M應(yīng)為大于0的數(shù),故結(jié)論錯(cuò)誤.4.根據(jù)對(duì)數(shù)的定義,因?yàn)?-2)4=16,所以log(-2)16=4.

(

?)對(duì)數(shù)的底數(shù)a應(yīng)滿(mǎn)足a>0且a≠1.x2=

(x>0,且x≠1).

(√)由對(duì)數(shù)的換底公式得logx2=

=

,故結(jié)論正確.2(-2a+1)有意義的a的取值范圍是

.

(√)要使對(duì)數(shù)log2(-2a+1)有意義,必須使-2a+1>0,解得a<

,故結(jié)論正確.“改變世界面貌的十個(gè)數(shù)學(xué)公式”被寫(xiě)到郵票中,第4枚是納皮爾指數(shù)與對(duì)數(shù)關(guān)

系公式:elnN=N,其中e=2.71828….伽利略曾發(fā)出豪言壯語(yǔ):“給我時(shí)間、空間和對(duì)

數(shù),我可以創(chuàng)造出一個(gè)宇宙來(lái).”對(duì)數(shù)性質(zhì)與對(duì)數(shù)運(yùn)算的應(yīng)用問(wèn)題ln4+1?提示:eln4+1=eln4·e=4e.aN=x和指數(shù)式ax=N推出

=N(a>0,且a≠1,N>0)?提示:把x=logaN代入ax=N,得

=N.3.在指數(shù)與對(duì)數(shù)的互化中,要注意什么?提示:要注意底數(shù)的范圍,如(-2)2=4,不能寫(xiě)成log(-2)4=2,只有a>0,a≠1,N>0時(shí),才有ax

=N?x=logaN.利用對(duì)數(shù)的運(yùn)算性質(zhì)可以將同底對(duì)數(shù)式進(jìn)行恒等變形,解題時(shí)要注意下列問(wèn)題:(1)利用對(duì)數(shù)的運(yùn)算性質(zhì),可以把乘、除、乘方的對(duì)數(shù)運(yùn)算轉(zhuǎn)化為對(duì)數(shù)的加、

減、乘法運(yùn)算,反之亦然.因此在計(jì)算時(shí),要靈活運(yùn)用運(yùn)算性質(zhì).(2)在使用公式的過(guò)程中,要注意公式成立的條件.

計(jì)算下列各式:(1)log535-2log5

+log57-log5

;(2)lg25+

lg8+lg5×lg20+(lg2)2.解析

(1)原式=log5(5×7)-2(log57-log53)+log57-(log59-log55)=log55+log57-2log57+2

log53+log57-2log53+log55=2log55=2.(2)原式=2lg5+2lg2+(1-lg2)(1+lg2)+(lg2)2=2(lg5+lg2)+1-(lg2)2+(lg2)2=2+1=3.

對(duì)數(shù)換底公式及其應(yīng)用

對(duì)不同底的對(duì)數(shù)式進(jìn)行恒等變形時(shí),可以用對(duì)數(shù)換底公式將底數(shù)換為相同的

數(shù),可以根據(jù)題目的特點(diǎn)換為指定的底數(shù),也可以換為常用對(duì)數(shù)或自然對(duì)數(shù).

(2)已知log189=a,18b=5,用a、b表示log3645為

.解析

(1)原式=

+

×

=

log23×

=

.(2)解法一:∵18b=5,∴l(xiāng)og185=b,∴l(xiāng)og3645=

=

=

=

=

.解法二:∵log189=a,18b=5,∴l(xiāng)g9=alg18,lg5=blg18,∴l(xiāng)og3645=

=

=

=

=

.解法三:∵log189=a,∴18a=9.(1)化簡(jiǎn):(log43+log83)(log32+log92)=

;又∵18b=5,∴45=5×9=18b·18a=18a+b.令log3645=x,則36x=45=18a+b,即36x=

=18a+b,∴

=18a+b,∴xlog18

=a+b,∴x=

=

.答案(1)

(2)

解題模板用已知對(duì)數(shù)式表示未知對(duì)數(shù)式,此類(lèi)問(wèn)題的本質(zhì)是把目標(biāo)分解為基本“粒子”,

然后用指定字母換元.跟蹤訓(xùn)練2(

)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)撥思路一:設(shè)ax=by=cz=t(t>0,且t≠1),可得x=logat,y=logbt,z=logct,代入

+

+

=0,利用對(duì)數(shù)的運(yùn)算性質(zhì)即可求得abc的值.思路二:設(shè)ax=by=cz=t(t>0,且t≠1),可得x=

,y=

,z=

,代入

+

+

=0,利用對(duì)數(shù)的運(yùn)算性質(zhì)即可求得abc的值.解析解法一:設(shè)ax=by=cz=t(t>0,且t≠1),∴x=logat,y=log