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2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算高中數(shù)學(xué)人教A版必修一第二章基本初等函數(shù)(I)2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算高中數(shù)學(xué)人教A版必修1x=③式x是否存在呢?問(wèn)題導(dǎo)入①③②?x=③式x是否存在呢?問(wèn)題導(dǎo)入①③②?2指數(shù)函數(shù)圖像如右:所以指數(shù)式

中的X存在!問(wèn)題探究這是已知底數(shù)和冪的值,求指數(shù)的問(wèn)題。指數(shù)函數(shù)圖像如右:所以指數(shù)式3為了解決“已知底數(shù)和冪的值,求指數(shù)”這類問(wèn)題,

引進(jìn)對(duì)數(shù)為了解決“已知底數(shù)和冪的值,求指數(shù)”這類問(wèn)題,4一般地,如果ax=N(a>0,且a≠1),那么數(shù)x叫做以a為底N的對(duì)數(shù)(logarithm),記作其中a叫做對(duì)數(shù)的底數(shù),N叫做真數(shù)。對(duì)數(shù)定義Nalog對(duì)數(shù)的寫(xiě)法一般地,如果ax=N(a>0,且a≠1),對(duì)數(shù)定義Nalo5對(duì)數(shù)定義的剖析問(wèn)題1:的含義是什么?含義?對(duì)數(shù)是一個(gè)數(shù),對(duì)數(shù)者,指數(shù)也。對(duì)數(shù)定義的剖析問(wèn)題1:的含義是什么?6根據(jù)對(duì)數(shù)的定義,表示下列x?習(xí)得定義,初步應(yīng)用根據(jù)對(duì)數(shù)的定義,表示下列x?習(xí)得定義,初步應(yīng)用7

指數(shù)—對(duì)數(shù)

底數(shù)—底數(shù)

冪—真數(shù)問(wèn)題2:指數(shù)式與對(duì)數(shù)式中a,x,N分別叫什么?對(duì)數(shù)定義的剖析指數(shù)—對(duì)數(shù)底數(shù)—底數(shù)冪—真數(shù)問(wèn)題2:指數(shù)式與對(duì)數(shù)式8

指數(shù)—對(duì)數(shù)

底數(shù)—底數(shù)

冪—真數(shù)問(wèn)題3:在對(duì)數(shù)式中a,x

,N的取值范圍是什么?N>0R,(負(fù)數(shù)和零沒(méi)有對(duì)數(shù))兩個(gè)等式所表示的是a,x,N這3個(gè)量之間的同一個(gè)關(guān)系.兩種寫(xiě)法可以相互轉(zhuǎn)化.對(duì)數(shù)定義的剖析指數(shù)—對(duì)數(shù)底數(shù)—底數(shù)冪—真數(shù)問(wèn)題3:在對(duì)數(shù)式中a9指數(shù)式與對(duì)數(shù)式互化

(1)(2)(3)(4)

應(yīng)用舉例1指數(shù)式與對(duì)數(shù)式互化

(1)(2)(3)(4)應(yīng)用舉例110問(wèn)題1:的含義是什么?含義?(5)ln1=這是已知底數(shù)和冪的值,求指數(shù)的問(wèn)題。自然對(duì)數(shù)求下列各式中x的值:(3)lne一般地,如果ax=N(a>0,且a≠1),兩種寫(xiě)法可以相互轉(zhuǎn)化.問(wèn)題3:在對(duì)數(shù)式中a,x,N的取值范圍是什么?這是已知底數(shù)和冪的值,求指數(shù)的問(wèn)題。(6)lg10=兩種寫(xiě)法可以相互轉(zhuǎn)化.第二章基本初等函數(shù)(I)那么數(shù)x叫做以a為底N的對(duì)數(shù)(logarithm),(6)lg10=log31=根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:(5)ln1=

常用對(duì)數(shù)

以10為底的對(duì)數(shù)稱為常用對(duì)數(shù),即簡(jiǎn)記為lgN

兩種特殊的對(duì)數(shù)問(wèn)題1:的含義是什么?含義?以11

自然對(duì)數(shù)

以e為底的對(duì)數(shù)稱為自然對(duì)數(shù),把簡(jiǎn)記為lnN

自然常數(shù)

兩種特殊的對(duì)數(shù)以e為底的對(duì)數(shù)稱為自然對(duì)數(shù),兩種特殊的對(duì)數(shù)12根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:

應(yīng)用探究log31=

(6)lg10=(1)log33=(2)(4)(5)

ln1=(4)(3);;;;;.log0.50.5=根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:應(yīng)用探究log31=13根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:

應(yīng)用探究log31=

(6)lg10=(1)log33=(2)(4)(5)

ln1=(4)(3);;;;;.log0.50.5=000111觀察上述各式,進(jìn)行適當(dāng)分類,歸納出一般性結(jié)論.根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:應(yīng)用探究log31=14=0

應(yīng)用探究1log31(1)(4)■=0■■(5)

ln1=0

對(duì)任意a>0且a≠1,都有(1的對(duì)數(shù)為0)結(jié)論:=0應(yīng)用探究1log31(1)(4)■=0■■15

應(yīng)用探究2=1=1=1(6)lg10log33(2)■■(3)

lne■

對(duì)任意a>0且a≠1,都有(底數(shù)的對(duì)數(shù)為1)

結(jié)論:應(yīng)用探究2=1=1=1(6)lg1016求下列各式中x的值:

(1)(2)解:(1)因?yàn)樗?2)因?yàn)樗杂谑?/p>

應(yīng)用舉例2求下列各式中x的值:(1)(2)解:(1)因?yàn)樗?217解:(1)(4)(3)(2)求下列各式的值(1)(4)(3)(2)25log5

課堂練習(xí)解:(1)(4)(3)(2)求下列各式的值(1)(181.對(duì)數(shù)的定義2.掌握指數(shù)式與對(duì)數(shù)式的互化4.會(huì)用指數(shù)運(yùn)算求簡(jiǎn)單的對(duì)數(shù)值(a>0,a≠1)3.對(duì)數(shù)的基本性質(zhì)1.對(duì)數(shù)的定義2.掌握指數(shù)式與對(duì)數(shù)式的互化4.會(huì)用指數(shù)運(yùn)算求19關(guān)于對(duì)數(shù)的概念要注意以下幾點(diǎn)知識(shí)詳解關(guān)于對(duì)數(shù)的概念要注意以下幾點(diǎn)知識(shí)詳解20

2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算高中數(shù)學(xué)人教A版必修一第二章基本初等函數(shù)(I)2.2.1對(duì)數(shù)與對(duì)數(shù)運(yùn)算高中數(shù)學(xué)人教A版必修21x=③式x是否存在呢?問(wèn)題導(dǎo)入①③②?x=③式x是否存在呢?問(wèn)題導(dǎo)入①③②?22指數(shù)函數(shù)圖像如右:所以指數(shù)式

中的X存在!問(wèn)題探究這是已知底數(shù)和冪的值,求指數(shù)的問(wèn)題。指數(shù)函數(shù)圖像如右:所以指數(shù)式23為了解決“已知底數(shù)和冪的值,求指數(shù)”這類問(wèn)題,

引進(jìn)對(duì)數(shù)為了解決“已知底數(shù)和冪的值,求指數(shù)”這類問(wèn)題,24一般地,如果ax=N(a>0,且a≠1),那么數(shù)x叫做以a為底N的對(duì)數(shù)(logarithm),記作其中a叫做對(duì)數(shù)的底數(shù),N叫做真數(shù)。對(duì)數(shù)定義Nalog對(duì)數(shù)的寫(xiě)法一般地,如果ax=N(a>0,且a≠1),對(duì)數(shù)定義Nalo25對(duì)數(shù)定義的剖析問(wèn)題1:的含義是什么?含義?對(duì)數(shù)是一個(gè)數(shù),對(duì)數(shù)者,指數(shù)也。對(duì)數(shù)定義的剖析問(wèn)題1:的含義是什么?26根據(jù)對(duì)數(shù)的定義,表示下列x?習(xí)得定義,初步應(yīng)用根據(jù)對(duì)數(shù)的定義,表示下列x?習(xí)得定義,初步應(yīng)用27

指數(shù)—對(duì)數(shù)

底數(shù)—底數(shù)

冪—真數(shù)問(wèn)題2:指數(shù)式與對(duì)數(shù)式中a,x,N分別叫什么?對(duì)數(shù)定義的剖析指數(shù)—對(duì)數(shù)底數(shù)—底數(shù)冪—真數(shù)問(wèn)題2:指數(shù)式與對(duì)數(shù)式28

指數(shù)—對(duì)數(shù)

底數(shù)—底數(shù)

冪—真數(shù)問(wèn)題3:在對(duì)數(shù)式中a,x

,N的取值范圍是什么?N>0R,(負(fù)數(shù)和零沒(méi)有對(duì)數(shù))兩個(gè)等式所表示的是a,x,N這3個(gè)量之間的同一個(gè)關(guān)系.兩種寫(xiě)法可以相互轉(zhuǎn)化.對(duì)數(shù)定義的剖析指數(shù)—對(duì)數(shù)底數(shù)—底數(shù)冪—真數(shù)問(wèn)題3:在對(duì)數(shù)式中a29指數(shù)式與對(duì)數(shù)式互化

(1)(2)(3)(4)

應(yīng)用舉例1指數(shù)式與對(duì)數(shù)式互化

(1)(2)(3)(4)應(yīng)用舉例130問(wèn)題1:的含義是什么?含義?(5)ln1=這是已知底數(shù)和冪的值,求指數(shù)的問(wèn)題。自然對(duì)數(shù)求下列各式中x的值:(3)lne一般地,如果ax=N(a>0,且a≠1),兩種寫(xiě)法可以相互轉(zhuǎn)化.問(wèn)題3:在對(duì)數(shù)式中a,x,N的取值范圍是什么?這是已知底數(shù)和冪的值,求指數(shù)的問(wèn)題。(6)lg10=兩種寫(xiě)法可以相互轉(zhuǎn)化.第二章基本初等函數(shù)(I)那么數(shù)x叫做以a為底N的對(duì)數(shù)(logarithm),(6)lg10=log31=根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:(5)ln1=

常用對(duì)數(shù)

以10為底的對(duì)數(shù)稱為常用對(duì)數(shù),即簡(jiǎn)記為lgN

兩種特殊的對(duì)數(shù)問(wèn)題1:的含義是什么?含義?以31

自然對(duì)數(shù)

以e為底的對(duì)數(shù)稱為自然對(duì)數(shù),把簡(jiǎn)記為lnN

自然常數(shù)

兩種特殊的對(duì)數(shù)以e為底的對(duì)數(shù)稱為自然對(duì)數(shù),兩種特殊的對(duì)數(shù)32根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:

應(yīng)用探究log31=

(6)lg10=(1)log33=(2)(4)(5)

ln1=(4)(3);;;;;.log0.50.5=根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:應(yīng)用探究log31=33根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:

應(yīng)用探究log31=

(6)lg10=(1)log33=(2)(4)(5)

ln1=(4)(3);;;;;.log0.50.5=000111觀察上述各式,進(jìn)行適當(dāng)分類,歸納出一般性結(jié)論.根據(jù)對(duì)數(shù)的定義,寫(xiě)出下列各對(duì)數(shù)的值:應(yīng)用探究log31=34=0

應(yīng)用探究1log31(1)(4)■=0■■(5)

ln1=0

對(duì)任意a>0且a≠1,都有(1的對(duì)數(shù)為0)結(jié)論:=0應(yīng)用探究1log31(1)(4)■=0■■35

應(yīng)用探究2=1=1=1(6)lg10log33(2)■■(3)

lne■

對(duì)任意a>0且a≠1,都有(底數(shù)的對(duì)數(shù)為1)

結(jié)論:應(yīng)用探究2=1=1=1(6)lg1036求下列各式中x的值:

(1)(2)解:(1)因?yàn)樗?2)因?yàn)樗杂谑?/p>

應(yīng)用舉例2求下列各式中x的值:(1)(2)解:(1)因?yàn)樗?237解:(

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