3.23.2對(duì)數(shù)與對(duì)數(shù)函數(shù)對(duì)數(shù)與對(duì)數(shù)函數(shù)3.2.13.2.1對(duì)數(shù)及其運(yùn)算對(duì)數(shù)及其運(yùn)算第第1 1課時(shí)對(duì)數(shù)的概念、常用對(duì)數(shù)課時(shí)對(duì)數(shù)的概念、常用對(duì)數(shù)目標(biāo)導(dǎo)航目標(biāo)導(dǎo)航課標(biāo)要求課標(biāo)要求1.1.理解對(duì)數(shù)的概念理解對(duì)數(shù)的概念, ,能進(jìn)行指數(shù)式與對(duì)數(shù)式的互化能進(jìn)行指數(shù)式與對(duì)數(shù)式的互化. .2.2.理解對(duì)數(shù)的底數(shù)和真數(shù)的范圍理解對(duì)數(shù)的底數(shù)和真數(shù)的范圍, ,掌握對(duì)數(shù)的基本性質(zhì)掌握對(duì)數(shù)的基本性質(zhì)及對(duì)數(shù)恒等式及對(duì)數(shù)恒等式. .素養(yǎng)達(dá)成素養(yǎng)達(dá)成通過對(duì)數(shù)概念的學(xué)習(xí)通過對(duì)數(shù)概念的學(xué)習(xí), ,培養(yǎng)數(shù)學(xué)抽象與數(shù)學(xué)運(yùn)算的核心培養(yǎng)數(shù)學(xué)抽象與數(shù)學(xué)運(yùn)算的核心素養(yǎng)素養(yǎng). .新知探求新知探求課堂探究課堂探究新知探求新知探求素養(yǎng)養(yǎng)成素養(yǎng)養(yǎng)成點(diǎn)擊進(jìn)入點(diǎn)擊進(jìn)入 情境導(dǎo)學(xué)情境導(dǎo)學(xué)知識(shí)探究知識(shí)探究1.1.對(duì)數(shù)的概念對(duì)數(shù)的概念指數(shù)函數(shù)指數(shù)函數(shù)y y= =a ax x(a(a00且且a1),a1),那么那么 叫做以叫做以 為底為底 的對(duì)數(shù)的對(duì)數(shù), ,記作記作x x= =logloga ay,y,讀作讀作x x等于等于 . .一般地一般地, ,對(duì)于指數(shù)式對(duì)于指數(shù)式a ab b= =N,N,有有b b= =logloga aN(aN(a0,0,且且a1),a1),其中其中, ,數(shù)數(shù)a a叫做叫做 ,N,N叫做叫做 . .冪指數(shù)冪指數(shù)x xa ay y以以a a為底為底y y的對(duì)數(shù)的對(duì)數(shù)對(duì)數(shù)的底數(shù)對(duì)數(shù)的底數(shù)真數(shù)真數(shù)2.2.對(duì)數(shù)恒等式是對(duì)數(shù)恒等式是 (a0(a0且且a1).a1).3.3.對(duì)數(shù)對(duì)數(shù)logloga aN(aN(a00且且a1)a1)的性質(zhì)的性質(zhì)(1)(1) 沒有對(duì)數(shù)沒有對(duì)數(shù), ,即即 ; ;(2)(2) 的對(duì)數(shù)為的對(duì)數(shù)為0,0,即即 ; ;(3)(3) 的對(duì)數(shù)等于的對(duì)數(shù)等于1,1,即即logloga aa a=1.=1.4.4.常用對(duì)數(shù)常用對(duì)數(shù)以以 為底的對(duì)數(shù)叫做常用對(duì)數(shù)為底的對(duì)數(shù)叫做常用對(duì)數(shù),log,log1010N N記作記作 . .0 0和負(fù)數(shù)和負(fù)數(shù)N0N01 1logloga a1=01=0底數(shù)底數(shù)1010lglg N N1.1.(2018(2018甘肅蘭州五十三中期中甘肅蘭州五十三中期中) )如果如果N=aN=a2 2(a0(a0且且a1),a1),則有則有( ( ) )(A)log(A)log2 2N=aN=a (B)log(B)log2 2a=Na=N( (C)logC)logN Na a=2=2 ( (D)logD)loga aN N=2=2自我檢測(cè)自我檢測(cè)D D解析解析: :因?yàn)橐驗(yàn)镹=aN=a2 2(a0(a0且且a1),a1),所以所以2=2=logloga aN N, ,故選故選D.D.A AB B類型一類型一 指數(shù)式、對(duì)數(shù)式互化指數(shù)式、對(duì)數(shù)式互化課堂探究課堂探究素養(yǎng)提升素養(yǎng)提升思路點(diǎn)撥思路點(diǎn)撥: :利用指數(shù)式與對(duì)數(shù)式的互化公式利用指數(shù)式與對(duì)數(shù)式的互化公式a ab b= =N Nb b= =logloga aN N來完成來完成. .解解: :(1)(1)因?yàn)橐驗(yàn)? 54 4=625,=625,所以所以loglog5 5625=4.625=4.方法技巧方法技巧 并非任何指數(shù)式都可以直接化為對(duì)數(shù)式并非任何指數(shù)式都可以直接化為對(duì)數(shù)式, ,如如(-3)(-3)2 2=9=9就不能直就不能直接寫成接寫成loglog-3-39=2,9=2,只有符合只有符合a0,a1a0,a1且且N0N0時(shí)時(shí), ,才有才有a ax x= =N Nx x= =logloga aN N. .變式訓(xùn)練變式訓(xùn)練1 1- -1:1:(1)(1)若若loglog5 5x=2,x=2,則則x=x=; ; (2)(2)若若logloga a2=m,log2=m,loga a3=n,3=n,則則a a2m+n2m+n= =. . 解析解析: :(1)(1)由指數(shù)式與對(duì)數(shù)式互化公式得由指數(shù)式與對(duì)數(shù)式互化公式得x=5x=52 2=25.=25.(2)(2)因?yàn)橐驗(yàn)閘ogloga a2=m,log2=m,loga a3=n,3=n,所以所以a am m=2,a=2,an n=3,=3,所以所以a a2m+n2m+n=(a=(am m) )2 2a an n=4=43=12.3=12.答案答案: :(1)25(1)25(2)12(2)12類型二類型二 對(duì)數(shù)基本性質(zhì)的應(yīng)用對(duì)數(shù)基本性質(zhì)的應(yīng)用【例例2 2】 求下列各式中求下列各式中x x的值的值: :(1)log(1)log3 3(x(x2 2-1)=0;-1)=0;(2)log(2)logx+3x+3(x(x2 2+3x)=1.+3x)=1.方法技巧方法技巧 有關(guān)有關(guān)“底數(shù)底數(shù)”和和“1 1”的對(duì)數(shù)的對(duì)數(shù), ,可利用對(duì)數(shù)的性質(zhì)知其值為可利用對(duì)數(shù)的性質(zhì)知其值為“1 1”和和“0 0”, ,化為常數(shù)化為常數(shù). .變式訓(xùn)練變式訓(xùn)練2 2- -1:1:求下列各式中求下列各式中x x的值的值: :(1)log(1)log2 2(log(log5 5x)=0;x)=0;(2)log(2)log3 3(lg x)=1.(lg x)=1.解解: :(1)(1)因?yàn)橐驗(yàn)閘oglog2 2(log(log5 5x)=0,x)=0,所以所以loglog5 5x=2x=20 0=1,=1,所以所以x=5x=51 1=5.=5.(2)(2)因?yàn)橐驗(yàn)閘oglog3 3(lg x)=1,(lg x)=1,所以所以lg x=3lg x=31 1=3,=3,所以所以x=10 x=103 3=1 000.=1 000.類型三類型三 由對(duì)數(shù)的定義及對(duì)數(shù)恒等式求值由對(duì)數(shù)的定義及對(duì)數(shù)恒等式求值方法技巧方法技巧 對(duì)數(shù)恒等式是利用對(duì)數(shù)定義推出的對(duì)數(shù)恒等式是利用對(duì)數(shù)定義推出的, ,要注意結(jié)構(gòu)特點(diǎn)要注意結(jié)構(gòu)特點(diǎn):(1):(1)它它們是同底的們是同底的;(2);(2)指數(shù)中含有對(duì)數(shù)形式指數(shù)中含有對(duì)數(shù)形式;(3);(3)其值為對(duì)數(shù)的真數(shù)其值為對(duì)數(shù)的真數(shù). .(2)(2)原式原式=10=10lg 9lg 91010lg 2lg 2=9=92=18.2=18.(3)(3)原式原式=bc.=bc.