第三章第五節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)

第五節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)課程標(biāo)準(zhǔn)1.理解對(duì)數(shù)的概念及其運(yùn)算性質(zhì),會(huì)用換底公式將一般對(duì)數(shù)轉(zhuǎn)化成自然對(duì)數(shù)或常用對(duì)數(shù).2.了解對(duì)數(shù)函數(shù)的概念.能畫出具體對(duì)數(shù)函數(shù)的圖象,了解對(duì)數(shù)函數(shù)的單調(diào)性與特殊點(diǎn).3.知道對(duì)數(shù)函數(shù)y=logax與指數(shù)函數(shù)y=ax(a>0,a≠1)互為反函數(shù).考情分析考點(diǎn)考法:高考命題常以考查對(duì)數(shù)的運(yùn)算性質(zhì)為主,考查學(xué)生的運(yùn)算能力;對(duì)數(shù)函數(shù)的單調(diào)性及應(yīng)用是考查熱點(diǎn),常以選擇題或填空題的形式出現(xiàn).核心素養(yǎng):數(shù)學(xué)抽象、邏輯推理、數(shù)學(xué)運(yùn)算.【必備知識(shí)·逐點(diǎn)夯實(shí)】【知識(shí)梳理·歸納】1.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a為底N的對(duì)數(shù),記作x=logaN,其中a叫做對(duì)數(shù)的底數(shù),N叫做真數(shù).以10為底的對(duì)數(shù)叫做常用對(duì)數(shù),記為lgN.

以e為底的對(duì)數(shù)叫做自然對(duì)數(shù),記為lnN.

2.對(duì)數(shù)的性質(zhì)與運(yùn)算性質(zhì)(1)對(duì)數(shù)的性質(zhì):loga1=0,logaa=1,alogaN=N(a>0,且a≠1,(2)對(duì)數(shù)的運(yùn)算性質(zhì)如果a>0,且a≠1,M>0,N>0,那么:①loga(MN)=logaM+logaN;②logaMN=logaMlogaN③logaMn=nlogaM(n∈R).(3)換底公式:logab=logcblogca(a>0,且a≠1,b【微點(diǎn)撥】(1)換底公式的變形①logab·logba=1,即logab=1logba(a,b均大于②logambn=nmlogab(a,b均大于0且不等于1,m≠0,n∈③logNM=logaMlogaN=logbMlogb(2)換底公式的推廣logab·logbc·logcd=logad(a,b,c均大于0且不等于1,d>0).3.對(duì)數(shù)函數(shù)的圖象與性質(zhì)y=logaxa>10<a<1圖象定義域(0,+∞)值域R性質(zhì)過定點(diǎn)(1,0),即x=1時(shí),y=0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,+∞)上是增函數(shù)在(0,+∞)上是減函數(shù)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)y=logax(a>0且a≠1)互為反函數(shù),它們的圖象關(guān)于直線y=x對(duì)稱.【基礎(chǔ)小題·自測(cè)】類型辨析改編易錯(cuò)高考題號(hào)12431.(多維辨析)(多選題)下列結(jié)論錯(cuò)誤的是 ()A.log2x2=2log2xB.若MN>0,則loga(MN)=logaM+logaNC.函數(shù)y=ln1+x1-x與y=ln(1+x)lnD.當(dāng)x>1時(shí),若logax>logbx,則a<b【解析】選ABD.Alog2x2=2log2|x|×B當(dāng)M<0,N<0時(shí),雖然MN>0,但loga(MN)=logaM+logaN不成立×D若0<b<1<a,則當(dāng)x>1時(shí),logax>logbx×2.(人A必修第一冊(cè)P141T13(1)·變形式)設(shè)a=log0.26,b=log0.36,c=log0.46,則 ()A.a>b>c B.a>c>bC.b>c>a D.c>b>a【解析】選A.方法一:如圖,作出函數(shù)y1=log0.2x,y2=log0.3x,y3=log0.4x的圖象,由圖可知,當(dāng)x=6時(shí),log0.26>log0.36>log0.46,即a>b>c.方法二:易知0>log60.4>log60.3>log60.2,所以1log60.即log0.46<log0.36<log0.26,即a>b>c.3.(2022·浙江高考)已知2a=5,log83=b,則4a3b= ()A.25 B.5 C.259 D.【解析】選C.由2a=5兩邊取以2為底的對(duì)數(shù),得a=log25.又b=log83=log23log28=13log23,所以a3b=log25log23=log253=log453log42=2log4534.(忽視對(duì)數(shù)函數(shù)的單調(diào)性)函數(shù)y=logax(a>0,a≠1)在[2,4]上的最大值與最小值的差是1,則a的值為________.

【解析】當(dāng)a>1時(shí),依題意得loga4loga2=1,解得a=2;當(dāng)0<a<1時(shí),依題意得loga2loga4=1,解得a=12答案:2或1【巧記結(jié)論·速算】1.對(duì)數(shù)函數(shù)y=logax(a>0且a≠1)的圖象恒過點(diǎn)(1,0),(a,1),(1a,1)2.如圖給出4個(gè)對(duì)數(shù)函數(shù)的圖象,則b>a>1>d>c>0,即在第一象限,不同的對(duì)數(shù)函數(shù)圖象從左到右底數(shù)逐漸增大.【即時(shí)練】1.函數(shù)y=loga(x2)+2(a>0且a≠1)的圖象恒過定點(diǎn)__________.

【解析】因?yàn)閘oga1=0,令x2=1,所以x=3,所以y=loga1+2=2,所以原函數(shù)的圖象恒過定點(diǎn)(3,2).答案:(3,2)2.已知圖中曲線C1,C2,C3,C4是函數(shù)y=logax(a>0,且a≠1)的圖象,則曲線C1,C2,C3,C4對(duì)應(yīng)的a的值依次為 ()A.3,2,13,12 B.2,3,1C.2,3,12,13 D.3,2,1【解析】選B.方法一:可用結(jié)論2,畫直線y=1,交點(diǎn)的位置自左向右,底數(shù)由小到大.方法二:因?yàn)镃1,C2為增函數(shù),所以它們的底數(shù)都大于1,又當(dāng)x>1時(shí),圖象越靠近x軸,其底數(shù)越大,故C1,C2對(duì)應(yīng)的a值分別為2,3.又因?yàn)镃3,C4為減函數(shù),所以它們的底數(shù)都大于0小于1,此時(shí)當(dāng)x>1時(shí),圖象越靠近x軸,其底數(shù)越小,所以C3,C4對(duì)應(yīng)的a分別為13,12.綜上可得C1,C2,C3,C4對(duì)應(yīng)的a值依次為2,3,13【核心考點(diǎn)·分類突破】考點(diǎn)一對(duì)數(shù)的運(yùn)算[例1](1)下列運(yùn)算正確的是 ()A.2log1510+log1B.log427×log258×log95=8C.lg2+lg50=10D.log(2+3)(23)(log22【解析】選D.對(duì)于A,2log1510+log150.25=log15(102×0.25)=log1552=2,A錯(cuò)誤;對(duì)于B,log427×log258×log95=lg33lg2lg100=2,C錯(cuò)誤;對(duì)于D,log(2+3)(23)(log22)2=1(12)2(2)計(jì)算:(1-lo【解析】原式=1=1=2(1-log6答案:1(3)已知a=log26,3b=36,則1a+2b=__________,2a【解析】a=log26,3b=36,則b=log336=2log36,則1a+2b=log62+log63=logab=log262log36=lg6lg22lg6則2ab=2lo答案:13【解題技法】對(duì)數(shù)運(yùn)算的一般思路(1)拆: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),然后再用對(duì)數(shù)的運(yùn)算性質(zhì)化簡(jiǎn)合并.(2)合: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)化為同底數(shù)對(duì)數(shù)真數(shù)的積、商、冪的運(yùn)算.【加練備選】1.已知a=lg2,10b=3,則log56= ()A.a+b1+a B.a+b1【解析】選B.因?yàn)閍=lg2,10b=3,所以b=lg3,log56=lg6lg5=lg2+lg31-2.(2023·豫北名校聯(lián)考)已知2a=7b=k,若2a+1b=1,則k的值為 (A.28 B.114 C.14 D.【解析】選A.因?yàn)?a=7b=k,所以a=log2k,b=log7k,所以1a=logk2,1b=logk7,所以2a+1b=2logk2+logk7=logk28=1,3.計(jì)算:(827)

-23+eln3+log1【解析】原式=(23)

3×(-23)+312log2(212答案:3考點(diǎn)二對(duì)數(shù)函數(shù)的圖象及應(yīng)用[例2](1)函數(shù)f(x)=2log4(1x)的大致圖象是 ()【解析】選C.方法一:函數(shù)f(x)=2log4(1x)的定義域?yàn)?∞,1),排除A,B;函數(shù)f(x)=2log4(1x)在定義域上單調(diào)遞減,排除D.方法二(特值法):分別取x=12及x=1驗(yàn)證即可(2)金榜原創(chuàng)·易錯(cuò)對(duì)對(duì)碰①當(dāng)x∈(0,14]時(shí),x<logax,則實(shí)數(shù)a的取值范圍為________②當(dāng)x∈(0,14]時(shí),方程x=logax有解,則實(shí)數(shù)a的取值范圍為________【解析】①若x<logax在x∈(0,14]時(shí)成立,則0<a<1,且y=x的圖象在y=logax圖象的下方則14<loga1所以0<a<1,a12即實(shí)數(shù)a的取值范圍是(116,1)答案:(116,1②構(gòu)造函數(shù)f(x)=x和g(x)=logax,當(dāng)a>1時(shí),不滿足條件;當(dāng)0<a<1時(shí),由①可知,只需兩圖象在(0,14]上有交點(diǎn)即可則f(14)≥g(14),即14≥loga14,得所以a的取值范圍為(0,116]答案:(0,116【解題技法】對(duì)數(shù)函數(shù)圖象的識(shí)別及應(yīng)用方法(1)在識(shí)別函數(shù)圖象時(shí),要善于利用已知函數(shù)的性質(zhì)、函數(shù)圖象上的特殊點(diǎn)(與坐標(biāo)軸的交點(diǎn)、最高點(diǎn)、最低點(diǎn)等)排除不符合要求的選項(xiàng).(2)一些對(duì)數(shù)型方程、不等式問題常轉(zhuǎn)化為相應(yīng)的函數(shù)圖象問題,利用數(shù)形結(jié)合法求解.【對(duì)點(diǎn)訓(xùn)練】1.(2023·東城區(qū)質(zhì)檢)函數(shù)y=logax與y=x+a在同一平面直角坐標(biāo)系中的圖象可能是 ()【解析】選A.當(dāng)a>1時(shí),函數(shù)y=logax的圖象為選項(xiàng)B,D中過點(diǎn)(1,0)的曲線,此時(shí)函數(shù)y=x+a的圖象與y軸的交點(diǎn)的縱坐標(biāo)a應(yīng)滿足a>1,選項(xiàng)B,D中的圖象都不符合要求;當(dāng)0<a<1時(shí),函數(shù)y=logax的圖象為選項(xiàng)A,C中過點(diǎn)(1,0)的曲線,此時(shí)函數(shù)y=x+a的圖象與y軸的交點(diǎn)的縱坐標(biāo)a應(yīng)滿足0<a<1,只有選項(xiàng)A中的圖象符合要求.2.已知函數(shù)f(x)=|log2x|,實(shí)數(shù)a,b滿足0<a<b,且f(a)=f(b),若f(x)在[a2,b]上的最大值為2,則1a+b=__________【解析】因?yàn)閒(x)=|log2x|,所以f(x)的圖象如圖所示,又f(a)=f(b)且0<a<b,所以0<a<1,b>1且ab=1,所以a2<a,由圖知,f(x)max=f(a2)=|log2a2|=2log2a=2,所以a=12,所以b=2,所以1a+b答案:4考點(diǎn)三對(duì)數(shù)函數(shù)的性質(zhì)及應(yīng)用【考情提示】對(duì)數(shù)函數(shù)的性質(zhì)及應(yīng)用是高考命題的熱點(diǎn),多以選擇題或填空題的形式呈現(xiàn),重點(diǎn)考查比較大小、解不等式等問題,難度中檔.角度1比較大小[例3](1)設(shè)a=20.1,b=ln52,c=log3910,則a,b,c的大小關(guān)系是 (A.b>c>a B.a>c>bC.b>a>c D.a>b>c【解析】選D.因?yàn)閍=20.1>20=1,0=ln1<b=ln52<lne=1,c=log3910<log31=0,所以a>b(2)設(shè)a,b,c均為正數(shù),且2a=log12a,12b=log12b,12cA.a<b<c B.c<b<aC.c<a<b D.b<a<c【解析】選A.因?yàn)閍,b,c均為正數(shù),將a,b,c分別看成是函數(shù)圖象的交點(diǎn)的橫坐標(biāo).在同一平面直角坐標(biāo)系內(nèi)分別畫出y=2x,y=12x,y=log2x,y=log1由圖可知a<b<c.角度2解對(duì)數(shù)不等式[例4]設(shè)函數(shù)f(x)=log2x,x>0,log12(-x),xA.(1,0)∪(0,1)B.(∞,1)∪(1,+∞)C.(1,0)∪(1,+∞)D.(∞,1)∪(0,1)【解析】選C.由題意可得a>0,log2a>-log角度3對(duì)數(shù)函數(shù)性質(zhì)的綜合應(yīng)用[例5](1)(2023·鄭州模擬)設(shè)函數(shù)f(x)=ln|x+3|+ln|x3|,則f(x) ()A.是偶函數(shù),且在(∞,3)上單調(diào)遞減B.是奇函數(shù),且在(3,3)上單調(diào)遞減C.是奇函數(shù),且在(3,+∞)上單調(diào)遞增D.是偶函數(shù),且在(3,3)上單調(diào)遞增【解析】選A.函數(shù)f(x)的定義域?yàn)閧x|x≠±3},f(x)=ln|x+3|+ln|x3|=ln|x29|,令g(x)=|x29|,則f(x)=lng(x),函數(shù)g(x)的單調(diào)區(qū)間由圖象(圖略)可知,當(dāng)x∈(∞,3),x∈(0,3)時(shí),g(x)單調(diào)遞減,當(dāng)x∈(3,0),x∈(3,+∞)時(shí),g(x)單調(diào)遞增,由復(fù)合函數(shù)單調(diào)性同增異減得單調(diào)區(qū)間.由f(x)=ln|(x)29|=ln|x29|=f(x)得f(x)為偶函數(shù).(2)(2023·武漢模擬)函數(shù)f(x)=loga(32ax)在區(qū)間[1,2]上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為()A.(0,1) B.(34,1C.(0,34) D.(【解析】選C.設(shè)u(x)=32ax(a>0且a≠1),則u(x)是減函數(shù),要使得函數(shù)f(x)=loga(32ax)在[1,2]上單調(diào)遞增,只需y=logau為減函數(shù),且滿足u(x)=32ax>0在x∈[1,2]上恒成立,所以0<a<1,u(x)min=u(2)=3(3)(2023·惠州模擬)若函數(shù)f(x)=loga(x2ax+12)(a>0,且a≠1)有最小值,則實(shí)數(shù)a的取值范圍是________【解析】令u(x)=x2ax+12=(xa2)2+12a24,則u(x欲使函數(shù)f(x)=loga(x2ax+12)有最小值則有a解得1<a<2,即實(shí)數(shù)a的取值范圍為(1,2).答案:(1,2)【解題技法】1.比較對(duì)數(shù)大小的類型及相應(yīng)方法2.求解對(duì)數(shù)不等式的兩種類型及方法類型方法logax>logab借助y=logax的單調(diào)性求解,如果a的取值不確定,那么需分a>1與0<a<1兩種情況討論logax>b需先將b化為以a為底的對(duì)數(shù)式的形式,再借助y=logax的單調(diào)性求解3.在