全國(guó)統(tǒng)考2024高考數(shù)學(xué)一輪復(fù)習(xí)第二章函數(shù)2.6對(duì)數(shù)與對(duì)數(shù)函數(shù)學(xué)案理含解析北師大版

2.6對(duì)數(shù)與對(duì)數(shù)函數(shù)必備學(xué)問(wèn)預(yù)案自診學(xué)問(wèn)梳理1.對(duì)數(shù)的概念(1)依據(jù)下圖的提示填寫(xiě)與對(duì)數(shù)有關(guān)的概念:(2)a的取值范圍.

2.對(duì)數(shù)的性質(zhì)與運(yùn)算法則(1)對(duì)數(shù)的運(yùn)算法則假如a>0,且a≠1,M>0,N>0,那么①loga(MN)=;

②logaMN=③logaMn=(n∈R).

(2)對(duì)數(shù)的性質(zhì):alogaN=N(a>0,且a≠(3)對(duì)數(shù)換底公式:logab=logcblogca(a>0,且a≠1;b>3.對(duì)數(shù)函數(shù)的圖像與性質(zhì)函數(shù)y=logax(a>0,且a≠1)a>10<a<1圖像性質(zhì)定義域:

值域:R過(guò)定點(diǎn)

當(dāng)x>1時(shí),y>0;當(dāng)0<x<1時(shí),y<0當(dāng)x>1時(shí),y<0;當(dāng)0<x<1時(shí),y>0在(0,+∞)內(nèi)是

在(0,+∞)內(nèi)是

4.反函數(shù)指數(shù)函數(shù)y=ax(a>0,且a≠1)與對(duì)數(shù)函數(shù)(a>0,且a≠1)互為反函數(shù),它們的圖像關(guān)于直線對(duì)稱(chēng).

1.對(duì)數(shù)的性質(zhì)(a>0,且a≠1,b>0)(1)loga1=0;考點(diǎn)自診1.推斷下列結(jié)論是否正確,正確的畫(huà)“√”,錯(cuò)誤的畫(huà)“×”.(1)log2x2=2log2x.()(2)函數(shù)y=log2x及y=log133x都是對(duì)數(shù)函數(shù).((3)當(dāng)x>1時(shí),若logax>logbx,則a<b.()(4)函數(shù)f(x)=lgx-2x+2與g(x)=lg(x-2)-lg(x+2)是同一個(gè)函數(shù).(5)對(duì)數(shù)函數(shù)y=logax(a>0,且a≠1)的圖像過(guò)定點(diǎn)(1,0),且過(guò)點(diǎn)(a,1),1a,-1.(2.(2024陜西西安中學(xué)八模,理3)已知x·log32=1,則4x=()A.4 B.6 C.4log33.(2024山東歷城二中模擬四,3)已知a=log1516,b=log13π3,c=3-A.b<a<c B.a<c<bC.c<b<a D.b<c<a4.若函數(shù)y'=a|x|(a>0,且a≠1)的值域?yàn)閧y'|0<y'≤1},則函數(shù)y=loga|x|的圖像大致是()5.(2024河北保定一模,理13)若2a=10,b=log510,則1a+1b關(guān)鍵實(shí)力學(xué)案突破考點(diǎn)對(duì)數(shù)式的化簡(jiǎn)與求值【例1】化簡(jiǎn)下列各式:(1)lg37+lg70-lg3-((2)log34273·log5思索對(duì)數(shù)運(yùn)算的一般思路是什么?解題心得對(duì)數(shù)運(yùn)算的一般思路(1)首先利用冪的運(yùn)算把底數(shù)或真數(shù)進(jìn)行變形,化成分?jǐn)?shù)指數(shù)冪的形式,使冪的底數(shù)最簡(jiǎn),然后正用對(duì)數(shù)運(yùn)算性質(zhì)化簡(jiǎn)合并.(2)將對(duì)數(shù)式化為同底數(shù)對(duì)數(shù)的和、差、倍數(shù)運(yùn)算,然后逆用對(duì)數(shù)的運(yùn)算性質(zhì),轉(zhuǎn)化為同底對(duì)數(shù)真數(shù)的積、商、冪的運(yùn)算.對(duì)點(diǎn)訓(xùn)練1(1)(2024全國(guó)1,文8)設(shè)alog34=2,則4-a=()A.116 B.19 C.18(2)(2024山東泰安一模,5)已知定義在R上的函數(shù)f(x)的周期為4,當(dāng)x∈[-2,2)時(shí),f(x)=13x-x-4,則f(-log36)+f(log354)=()A.32 B.32-logC.-12 D.23+log考點(diǎn)對(duì)數(shù)函數(shù)的圖像及其應(yīng)用【例2】(1)函數(shù)y=2log4(1-x)的圖像大致是()(2)已知當(dāng)0<x≤12時(shí),4x=logax有解,則實(shí)數(shù)a的取值范圍是.變式發(fā)散將本例(2)中的“4x=logax有解”改為“4x<logax”,則實(shí)數(shù)a的取值范圍為.

解題心得應(yīng)用對(duì)數(shù)型函數(shù)的圖像可求解的問(wèn)題:(1)對(duì)一些可通過(guò)平移、對(duì)稱(chēng)變換作出其圖像的對(duì)數(shù)型函數(shù),在求解其單調(diào)性(單調(diào)區(qū)間)、值域(最值)、零點(diǎn)時(shí),也常利用數(shù)形結(jié)合思想;(2)一些對(duì)數(shù)型方程、不等式問(wèn)題常轉(zhuǎn)化為相應(yīng)的函數(shù)圖像問(wèn)題,利用數(shù)形結(jié)合法求解.對(duì)點(diǎn)訓(xùn)練2(1)函數(shù)f(x)=loga|x|+1(0<a<1)的圖像大致是()(2)函數(shù)y=|log2x|-12x的零點(diǎn)個(gè)數(shù)是()A.0 B.1 C.2 D.3考點(diǎn)對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用(多考向探究)考向1比較含對(duì)數(shù)的函數(shù)值的大小【例3】(1)(2024全國(guó)3,文10)設(shè)a=log32,b=log53,c=23,則(A.a<c<b B.a<b<c C.b<c<a D.c<a<b(2)(2024河北滄州一模,理9)已知a=log0.30.5,b=log30.5,c=log0.50.9,則()A.ab<ac<a+b B.a+b<ab<acC.ac<ab<a+b D.ab<a+b<ac解題心得比較含對(duì)數(shù)的函數(shù)值的大小,首先應(yīng)確定對(duì)應(yīng)函數(shù)的單調(diào)性,然后比較含對(duì)數(shù)的自變量的大小,同底數(shù)的可借助函數(shù)的單調(diào)性;底數(shù)不同、真數(shù)相同的可以借助函數(shù)的圖像;底數(shù)、真數(shù)均不同的可借助中間值(0或1).對(duì)點(diǎn)訓(xùn)練3(1)(2024山西太原二模,理3)已知a=log52,b=log0.50.2,c=0.50.2,則()A.a<b<c B.a<c<bC.b<a<c D.c<a<b(2)(2024全國(guó)3,理12)已知55<84,134<85.設(shè)a=log53,b=log85,c=log138,則()A.a<b<c B.b<a<cC.b<c<a D.c<a<b考向2解含對(duì)數(shù)的函數(shù)不等式【例4】(1)已知函數(shù)f(x)是定義在R上的偶函數(shù),且在區(qū)間[0,+∞)內(nèi)遞增.若正實(shí)數(shù)a滿(mǎn)意f(log2a)+f(log12a)≤2f(1),則a的取值范圍是(A.[1,2] B.0,12 C.12,2 D.(0,2](2)設(shè)函數(shù)f(x)=log2x,x>0,log12(-A.(-1,0)∪(0,1) B.(-∞,-1)∪(1,+∞)C.(-1,0)∪(1,+∞) D.(-∞,-1)∪(0,1)解題心得解簡(jiǎn)潔對(duì)數(shù)不等式,先統(tǒng)一底數(shù),化為形如logaf(x)>logag(x)(a>0,且a≠1)的不等式,再借助y=logax的單調(diào)性求解,當(dāng)a>1時(shí),logaf(x)>logag(x)?f(x)>0,g(x)>0,f(x對(duì)點(diǎn)訓(xùn)練4(1)定義在R上的奇函數(shù)f(x),當(dāng)x∈(0,+∞)時(shí),f(x)=log2x,則不等式f(x)<-1的解集是.

(2)不等式log(x-3)(x-1)≥2的解集為.

考向3對(duì)數(shù)型函數(shù)的綜合問(wèn)題【例5】已知函數(shù)f(x)=loga(x+1)-loga(1-x),a>0,且a≠1.(1)求f(x)的定義域;(2)推斷f(x)的奇偶性,并予以證明;(3)當(dāng)a>1時(shí),求使f(x)>0的x的取值范圍.解題心得有關(guān)對(duì)數(shù)型函數(shù)的綜合問(wèn)題要留意三點(diǎn):一是定義域,全部問(wèn)題都必需在定義域內(nèi)探討;二是底數(shù)與1的大小關(guān)系;三是復(fù)合函數(shù)的構(gòu)成,即它是由哪些基本初等函數(shù)復(fù)合而成的.另外,解題時(shí)要留意數(shù)形結(jié)合、分類(lèi)探討、轉(zhuǎn)化與化歸思想的運(yùn)用.對(duì)點(diǎn)訓(xùn)練5(1)(2024山東濰坊一模,7)已知定義在R上的偶函數(shù)f(x)=2|x-m|-1,記a=f(-ln3),b=f(-log25),c=f(2m),則()A.a<b<c B.a<c<bC.c<a<b D.c<b<a(2)已知函數(shù)f(x)=loga(ax2-x+3)在[1,3]上遞增,則a的取值范圍是.

1.多個(gè)對(duì)數(shù)函數(shù)圖像比較底數(shù)大小的問(wèn)題,可通過(guò)圖像與直線y=1交點(diǎn)的橫坐標(biāo)進(jìn)行判定.2.探討對(duì)數(shù)型函數(shù)的圖像時(shí),一般從最基本的對(duì)數(shù)函數(shù)的圖像入手,通過(guò)平移、伸縮、對(duì)稱(chēng)變換得到.特殊地,要留意底數(shù)a>1和0<a<1的兩種不同狀況.有些困難的問(wèn)題,借助于函數(shù)圖像來(lái)解決,就變得簡(jiǎn)潔了,這是數(shù)形結(jié)合思想的重要體現(xiàn).3.利用對(duì)數(shù)函數(shù)單調(diào)性可解決比較大小、解不等式、求最值等問(wèn)題,其基本方法是“同底法”,即把不同底的對(duì)數(shù)式化為同底的對(duì)數(shù)式,然后依據(jù)單調(diào)性來(lái)解決.1.在運(yùn)算性質(zhì)logaMn=nlogaM中,要特殊留意M>0的條件,當(dāng)n∈N+,且n為偶數(shù)時(shí),在無(wú)M>0的條件下應(yīng)為logaMn=nloga|M|.2.解決與對(duì)數(shù)函數(shù)有關(guān)的問(wèn)題時(shí)需留意兩點(diǎn):(1)定義域優(yōu)先的原則.(2)要有分類(lèi)探討的意識(shí).2.6對(duì)數(shù)與對(duì)數(shù)函數(shù)必備學(xué)問(wèn)·預(yù)案自診學(xué)問(wèn)梳理1.(1)指數(shù)對(duì)數(shù)冪真數(shù)底數(shù)(2)a>0,且a≠12.(1)①logaM+logaN②logaM-logaN③nlogaM3.(0,+∞)(1,0)增函數(shù)減函數(shù)4.y=logaxy=x考點(diǎn)自診1.(1)×(2)×(3)×(4)×(5)√2.D∵x·log32=1,∴x=log23,∴4x=4log23.Da=log1516>log1515=1,b=log13π3<lo4.A函數(shù)y'=a|x|(a>0,且a≠1)的值域?yàn)閧y'|0<y'≤1},則0<a<1,由此可知y=loga|x|的圖像大致是A.5.1∵2a=10,∴a=log210,又b=log510,∴1a+1b=1log2關(guān)鍵實(shí)力·學(xué)案突破例1解(1)原式=lg37(lg3)2-=1-|lg3-1|=lg3.(2)原式=log33343·log5[10-(332)23-7log72]=34-1·log55對(duì)點(diǎn)訓(xùn)練1(1)B(2)A(1)因?yàn)閍log34=log34a=2,所以4a=32=9,所以4-a=14a=(2)由題意,f(log354)=f(log354-4)=flog323,∵當(dāng)x∈[-2,2)時(shí),f(x)=13x-x-4,且-log36∈[-2,2),log323∈[-2,2),∴f(-log36)+f(log354)=13

-log36-(-log36)-4+13

log323-log323-4=3log36+3log332+log36例2(1)C(2)0,22(1)函數(shù)y=2log4(1-x)的定義域?yàn)?-∞,1),解除A,B;又函數(shù)y=2log4(1-x)在定義域內(nèi)遞減,解除D.故選C.(2)構(gòu)造函數(shù)f(x)=4x和g(x)=logax.當(dāng)a>1時(shí)不滿(mǎn)意條件,當(dāng)0<a<1時(shí),畫(huà)出兩個(gè)函數(shù)在0,12上的大致圖像,如圖所示.可知,只需兩圖像在0,12上有交點(diǎn)即可,則f12≥g12,即2≥loga12,則0<a≤22,所以a的取值范圍為0,22.變式發(fā)散22,1設(shè)函數(shù)f(x)=4x和g(x)=logax,可知當(dāng)a>1時(shí)不滿(mǎn)意條件,當(dāng)0<a<1時(shí),f12<g12,即2<loga12,則a>22,所以a的取值范圍為22,1.對(duì)點(diǎn)訓(xùn)練2(1)A(2)C(1)由于函數(shù)f(x)=loga|x|+1(0<a<1)是偶函數(shù),故其圖像關(guān)于y軸對(duì)稱(chēng).當(dāng)x>0時(shí),f(x)=loga|x|+1(0<a<1)遞減;當(dāng)x<0時(shí),f(x)=loga|x|+1(0<a<1)遞增.再由圖像過(guò)點(diǎn)(1,1),(-1,1),可知應(yīng)選A.(2)函數(shù)y=|log2x|-12x的零點(diǎn)個(gè)數(shù)即為方程|log2x|=12x的實(shí)數(shù)根的個(gè)數(shù).在同一平面直角坐標(biāo)系內(nèi)作出函數(shù)y=|log2x|及y=12x的圖像(圖像略),不難得出兩個(gè)函數(shù)的圖像有2個(gè)交點(diǎn),故選C.例3(1)A(2)D(1)∵32a=32log32=log3223=log98<1,∵32b=32log53=log5233=log∴b>23.又c=23,∴a<c<b.(2)∵0=log0.31<log0.30.5<log0.30.3=1,即0<a<1;log30.5<log31=0,即b<0;0=log0.51<log0.50.9<log0.50.5=1,即0<c<1,∴ab<0,0<ac<1,即有ab<ac.∵1a+1b=log0.50.3+log0.53=log0.50.9=c,即0∴ab<a+b<0.綜上,ab<a+b<ac.故選D.對(duì)點(diǎn)訓(xùn)練3(1)B(2)A(1)∵log51<log52<log55,∴0<a<12,b=log0.50.2=log1215=log25>∵0.51<0.50.2<0.50,∴12<c<1,∴a<c<b,故選B(2)∵43a=43log53=log5334=log12581<1,∵43b=43log85=log8354=log512625>1,∵55<84,∴54b=54log85=log8455<1,∵134<85,∴54c=54log138=log13485>綜上,a<b<c.例4(1)C(2)C(1)因?yàn)閘og12a=-log2a,所以f(log2a)+f(log12a)=f(log2a)+f(-log2a)=2f(log2a),原不等式變?yōu)?f(log2a)≤2f(1),即f(log2a)≤f(1).又因?yàn)閒(x)是定義在R上的偶函數(shù),且在[0,+∞)內(nèi)遞增,所以|log2a|≤1,即-1≤log2a≤1,解得1(2)由題意可得a>0,log2a>-lo對(duì)點(diǎn)訓(xùn)練4(1)(-∞,-2)∪0,12(2){x|4<x≤5}(1)由已知條件可知,當(dāng)x∈(-∞,0)時(shí),f(x)=-log2(-x).當(dāng)x∈(0,+∞)時(shí),f(x)<-1,即為log2x<-1,解得0<x<12當(dāng)x∈(-∞,0)時(shí),f(x)<-1,即為-log2(-x)