指數(shù)函數(shù)和對(duì)數(shù)函數(shù)復(fù)習(xí)有詳細(xì)知識(shí)點(diǎn)和習(xí)題詳-解-補(bǔ)課

指數(shù)函數(shù)和對(duì)數(shù)函數(shù)復(fù)習(xí)(有詳細(xì)知識(shí)點(diǎn)和習(xí)題詳-解)--補(bǔ)課第六講指數(shù)函數(shù)和對(duì)數(shù)函數(shù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)都是基本初等函數(shù)，是高中必須掌握的，在高考中，主要是考查基礎(chǔ)知識(shí)。要求掌握擴(kuò)充后指數(shù)的運(yùn)算，對(duì)數(shù)的運(yùn)算，指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的圖像和性質(zhì)。一、指數(shù)的性質(zhì),一,整數(shù)指數(shù)冪n0,1(整數(shù)指數(shù)冪概念:？a,a,a,,aaa,,10(n,N)，，,,,,,n個(gè)a1,,naanN,,,0,，，nanmnmn，mmn2(整數(shù)指數(shù)冪的運(yùn)算性質(zhì):(1)(2)aaamnZ,,,,aamnZ,,,，，，，，，nnnababnZ,,,(3)，，，，nnnaa,,mnmnmn,,,,1nn其中aaaaa,,,,，(,,,,,abab，，,,nbb,,3(的次方根的概念an,一般地，如果一個(gè)數(shù)的次方等于，那么這個(gè)數(shù)叫做的次方根，naan，，n,1,n,Nn,即:若x,a，則叫做的次方根，xan，，n,1,n,N33,27,27,,327,3例如:27的3次方根，的3次方根，55,32,32,,232,232的5次方根，的5次方根(nnna,0aa,0a,0說(shuō)明:?若是奇數(shù)，則的次方根記作;若則，若則;a,onannna,0a,a?若是偶數(shù)，且則的正的次方根記作，的負(fù)的次方根，記作:;(例如:nanan4,16,,2,8,,228的平方根16的4次方根)na,0a?若是偶數(shù)，且則沒(méi)意義，即負(fù)數(shù)沒(méi)有偶次方根;nn,n00,??;，，？0,0n,1,n,Nnnna?式子叫根式，叫根指數(shù)，叫被開(kāi)方數(shù)。?(aa,na，，4(的次方根的性質(zhì)annna,a一般地，若是奇數(shù)，則;naa,0,nna,a,若n是偶數(shù)，則(,,aa,0,5(例題分析:7，40，7,40例(計(jì)算:22,(5，2)，(5,2),257，40，7,40解:,二,分?jǐn)?shù)指數(shù)冪10125310212453aaaa,,,0aaaa,,,01(分?jǐn)?shù)指數(shù)冪:，，，，即當(dāng)根式的被開(kāi)方數(shù)能被根指數(shù)整除時(shí)，根式可以寫(xiě)成分?jǐn)?shù)指數(shù)冪的形式;1nmmn冪的運(yùn)算性質(zhì)對(duì)分?jǐn)?shù)指數(shù)冪也適用，aa,，，34225524，34，,,,,253254334435aaaaaa,,,,a,0例如:若，則，，?(aa,aa,,,,,,,,,m,nmn規(guī)定:(1)正數(shù)的正分?jǐn)?shù)指數(shù)冪的意義是;aaamnNn,,,,0,,,1，，m11,,n(2)正數(shù)的負(fù)分?jǐn)?shù)指數(shù)冪的意義是(aamnNn,,,,,0,,,1，，mnmana2(分?jǐn)?shù)指數(shù)冪的運(yùn)算性質(zhì):整數(shù)指數(shù)冪的運(yùn)算性質(zhì)對(duì)于分?jǐn)?shù)指數(shù)冪也同樣適用，即:rsrs，10,,aaaarsQ,,,，，，，srrs20,,aaarsQ,,,，，，，，，rrr30,0,abababrQ,,,,，，，，，，說(shuō)明:(1)有理數(shù)指數(shù)冪的運(yùn)算性質(zhì)對(duì)無(wú)理數(shù)指數(shù)冪同樣適用;(2)0的正分?jǐn)?shù)指數(shù)冪等于0，0的負(fù)分?jǐn)?shù)指數(shù)冪沒(méi)意義。3(例題分析:【例1】用分?jǐn)?shù)指數(shù)冪的形式表示下列各式ao,:，，2332aa,，，.aaaa,115，222222aa,aaaa,,,解:=;211332333aaa,,aa,=;1113322,,,,224aaaaaa,,,=(,,,,,,,,【例2】計(jì)算下列各式的值(式中字母都是正數(shù))(831211511,,,,,,,,,84336622mn263ababab,,,(1);(2);,,,,,,,,,,,,,,,,88833112115112,,,,,,,,,,,,,,m,238844336622mnmn263ababab,,,,mn解(1)(2)==(,,,,,,,,,,,,3n,,,,,,,,,,,,211115，,，,326236=263，,,,ab，，，，，，，，044aba,=;例3(計(jì)算下列各式:2a3451255,,a,0(1)(2)(，，，，32aa2312131522,,aa,,5555,,,51255,,,,aa解:(1)==(2)=(,,，，2132,,aa32aa55512412455,555,==;21133,,,12222【例3】已知，求下列各式的值:(1);(2).xx，xx，xx，,3111111,,,222222222解:(1)？()xx，,，，()2()xxxx11,,，,325，,，，xx211,22?xx，,,5，11,,122x,0又由得，?，xx，,0xx，,311,22所以xx，,5.3311111111,,,,,33222222222222(2)(法一)xx，,(xx)()，,，,，()[()()]xxxxxx11,,122,25，,，，,()[()1]xxxx,,5(31)333333,,,,33,222222222(法二),，，xx2[()()]xx，,，，()()2xxxx33,,,122而xx，,，，,()(1)xxxx,,1122,18,，，,()[()3]xxxx,，,3(33)33,222?，()20xx，,33,,122x,0xx，,0又由xx，,,30得，?，33,22xx，,,2025所以.二、指數(shù)函數(shù)1(指數(shù)函數(shù)定義:xa,0a,1R一般地，函數(shù)(且)叫做指數(shù)函數(shù)，其中是自變量，叫底數(shù)，函數(shù)定義域是(xaya,xa,101,,a2(指數(shù)函數(shù)在底數(shù)及這兩種情況下的圖象和性質(zhì):ya,a,101,,a圖象R(1)定義域:(2)值域:(0,)，,性x,0質(zhì)(0,1)y,1(3)過(guò)定點(diǎn)，即時(shí)RR(4)在上是增函數(shù)(4)在上是減函數(shù)3【例1】求下列函數(shù)的定義域、值域:1x1a,1x,xx,21(1)(2)(3)(4)(y,,1()yaa,,,(0,1)y,8y,3x2a，111？210x,,解:(1)?原函數(shù)的定義域是，x,{,}xxRx,,221令則t,ttR,,0,21x,t?得，yy,,0,1ytRt,,,8(,0)所以，原函數(shù)的值域是({0,1}yyy,,1xx,0(2)?原函數(shù)的定義域是，？,,0,，,1()0，,21x01,,t令t,,1()則，(0)x,2在是增函數(shù)?，0,101,,y？yt,，,所以，原函數(shù)的值域是0,1(，,R(3)原函數(shù)的定義域是，t,0令tx,,則，t在,,,0是增函數(shù)，?，01,,y？y,3，,所以，原函數(shù)的值域是0,1(，,R(4)原函數(shù)的定義域是，xa,1y，1x由yaa,,,(0,1)得，a,,xa，1y,1y，1x？a,0?，?，,,0,,,11yy,1,1,1所以，原函數(shù)的值域是(，，說(shuō)明:求復(fù)合函數(shù)的值域通過(guò)換元可轉(zhuǎn)換為求簡(jiǎn)單函數(shù)的值域。xa，1a,1y,【例2】當(dāng)時(shí)，證明函數(shù)是奇函數(shù)。xa,1xx,0a,,10證明:由得，，{0}xx,故函數(shù)定義域關(guān)于原點(diǎn)對(duì)稱。,xx,xxa，11，a(1)aa，fx(),,,,,fx(),,xx,xxa,11,a(1)aa,?fxfx()(),,,xa，1y,所以，函數(shù)是奇函數(shù)。xa,14三、對(duì)數(shù)的性質(zhì)bb1(對(duì)數(shù)定義:一般地，如果()的次冪等于N,就是，那么數(shù)b叫做a為底Na,0且a,1a,Nab的對(duì)數(shù)，記作，a叫做對(duì)數(shù)的底數(shù)，N叫做真數(shù)。即，。aN,logN,blogNb,aaaNbb底數(shù)冪指數(shù)指數(shù)式a,N對(duì)數(shù)式logN,b對(duì)數(shù)的底數(shù)真數(shù)對(duì)數(shù)a說(shuō)明:1(在指數(shù)式中冪N>0，?在對(duì)數(shù)式中，真數(shù)N>0((負(fù)數(shù)與零沒(méi)有對(duì)數(shù))？0a,0a,12(對(duì)任意且,都有?，同樣:(a,1log10,log1a,？aalogNbab3(如果把中的寫(xiě)成，則有(對(duì)數(shù)恒等式)(aN,aN,logNa2(對(duì)數(shù)式與指數(shù)式的互換22例如:，;，;416,10100,log162,log1002,4101,212，;，。42,100.01,log0.012,,log2,1042【例1】將下列指數(shù)式寫(xiě)成對(duì)數(shù)式:m4a1,,1,6(1)525,;(2);(3);(4)(327,,5.372,,,643,,1);(2);(3);(4)(解:(1log6254,log27,alog5.37,mlog6,,15326433(介紹兩種常見(jiàn)的對(duì)數(shù):?常用對(duì)數(shù):以10作底簡(jiǎn)寫(xiě)成;logNlgN10lnN?自然對(duì)數(shù):以作底為無(wú)理數(shù)，=2.71828……，簡(jiǎn)寫(xiě)成(logNeeelog625【例2】(1)計(jì)算:，(log2793453x23xx,927,33,解:設(shè)則，,?;log27x,924xx4343x,555,log6255625,令，?,,?(x,，，34532log3211xx，,,logx,,(2)求x的值:?;?(，，3,,221x,,,4,,3,14解:?;,,x3427222?32121200,2xxxxxxx，,,,,，,,,,,2,210x,,,2x,0x,,2但必須:，?舍去，從而(211x,,,2,3210xx，,,,37,,,log3log2(3)求底數(shù):?，?(xx585353,,,,3535x,3解:?x,,3(3)?;57788,,87x,2?,?(x,,22,,,,,,4(對(duì)數(shù)的運(yùn)算性質(zhì):如果a>0，a,1，M>0，N>0，那么(1);log()loglogMNMN,，aaaM(2);loglog-log,MNaaaNn(3)(loglog()MnMnR,,aa【例3】計(jì)算:7lg27，lg8,3lg10lg243(1)lg1421g，lg7,lg18;(2);(3)(,3lg9lg1.27解:(1)解法一:lg14,2lg，lg7,lg1832,，,,，,，lg(27)2(lg7lg3)lg7lg(32);,，,，，,,,lg2lg72lg72lg3lg72lg3lg207解法二:lg14,2lg，lg7,lg18372,,，,lg14lg()lg7lg18314，7lg=;,,lg1072()，1835lg243lg35lg35(2);,,,2lg92lg32lg3113(lg32lg21)，,3322lg27，lg8,3lg10lg(3)lg23lg103，,2(3)=(,,2lg1.232，lg32lg212，,lg10logNm5(換底公式:(a>0,a,1;)mm,,0,1logN,alogamxaN,logNx,證明:設(shè)，則，axxaNloglog,兩邊取以m為底的對(duì)數(shù)得:，?，loglogaN,mmmmlogNlogNmmx,logN,從而得:，?(alogalogamm說(shuō)明:兩個(gè)較為常用的推論:nnb,0loglogbb,loglog1ba，,a(1);(2)(、且均不為1)(maabamlgblgalogb,loga,,,1證明:(1);ablgalgb6nlglgbnbnn(2)(loglogbb,,,mamalglgamam1log3,40.2【例4】計(jì)算:(1);(2)(5log3log2log32,，492555解:(1)原式=;,,,151log30.21log55353115153log3log2log2(2)原式=(,，,，,232224442b【例5】已知，，求(用a,b表示)(185,log9,alog45183618解:?，?，log,1,log2,alog9,a1818182?，log21,,a18b又?，185,?，log5,b18log45log9，log5a，b181818log45,,,?(36log361，log22,a1818111xyz【例6】設(shè)3,4,6,t,1，求證:(,,zx2yxyz證明:?3,4,6,t,1，lgtlgtlgt?，x,，y,，z,lg3lg4lg611lg6lg3lg2lg41?,,,,,,(zxlgtlgtlgt2lgt2y四、對(duì)數(shù)函數(shù)1(對(duì)數(shù)函數(shù)的定義:函數(shù)y,logx叫做對(duì)數(shù)函數(shù)。(a,0且a,1)a2(對(duì)數(shù)函數(shù)的性質(zhì):y,logx(1)定義域、值域:對(duì)數(shù)函數(shù)的定義域?yàn)?，值域?yàn)?(a,0且a,1)(0,，,)(,,,，,)a(2)圖象:由于對(duì)數(shù)函數(shù)是指數(shù)函數(shù)的反函數(shù)，所以對(duì)數(shù)函數(shù)的圖象只須由相應(yīng)的指數(shù)函數(shù)圖象作關(guān)于y,x的對(duì)稱圖形，即可獲得。a,10,a,1y,logx同樣:也分與兩種情況歸納，以y,logx(圖1)與(圖2)為例。1221xxy,()y,2yx,2yx,11yx,log121yx,log12(圖2)(圖1)7(3)對(duì)數(shù)函數(shù)性質(zhì)列表:a,101,,ax,1x,1yx,loga圖象(1,0)(1,0)yx,loga(1)定義域:(0,)，,(2)值域:R性x,1(3)過(guò)點(diǎn)，即當(dāng)時(shí)，(1,0)y,0質(zhì)(4)在上是減函數(shù)(0,)，,(4)在(0，+?)上是增函數(shù)【例1】求下列函數(shù)的定義域:22(1);(2);(3)(y,log(4,x)y,logxy,log(9,x)aaa分析:此題主要利用對(duì)數(shù)函數(shù)的定義域求解。y,logx(0,)，,a2x,0解:(1)由>0得，x2?函數(shù)的定義域是;xx,0y,logx,,a4,x,0x,4(2)由得，?函數(shù)的定義域是;xx,4y,log(4,x),,a2,x,3,x,0(3)由9-得-3，2?函數(shù)的定義域是(xx,,,33y,log(9,x),,a【例2】比較下列比較下列各組數(shù)中兩個(gè)值的大小:0.91.1(1)，，;(2)，，(log0.9log0.8log3log3log31.10.75670.901.11.11,,解:(1)?，log0.9log10,,，1.11.10log1log0.8log0.71,,,,，0.70.70.70.91.1,?log0.8,log0.9(0.71.1(2)?0log5log6log7,,,，333log3,log3log3?(,567【例3】求下列函數(shù)的值域:2(1)yx,，log(3);(2)yx,,log(3)22tx,，3解:(1)令，則yt,log，2t,0R?，?，即函數(shù)值域?yàn)?yR,203,,ttx,,3(2)令，則，y,log3(,log3],,?，即函數(shù)值域?yàn)?222【例4】判斷函數(shù)的奇偶性。fxxx()log(1),，,22xx，,1解:?fx()(,),,，,恒成立，故的定義域?yàn)椋?fxxx()log(1),,，，281,,log22xx，，12xx，,1,,log2222(1)xx，,2，,,，,,,log1()xxfx2所以，為奇函數(shù)。fx()2【例5】求函數(shù)的單調(diào)區(qū)間。yxx,,，2log(32)13313322解:令在上遞增，在上遞減，[,)，,(,],,uxxx,,，,,,32()22242x,2x,1又?，?或，xx,，,3202故在上遞增，在上遞減，又?為減函數(shù)，uxx,,，32yu,2log(2,)，,(,1),,132所以，函數(shù)在上遞增，在上遞減。(2,)，,(,1),,yxx,,，2log(32)139課堂練習(xí)題(1)1、填空:mm31，35(3);(4);xx,()(),,,yy2343(5);(6);()()abab,,,(2)(2)(2),,,,m23n262、(1)若，則;(2)若，則;aaa,aaa,m,n,mnmn，23mn，(3)若，，用表示，;3,a3,3,3,bab,(2)m243(3);(4);()a,,,()x323，，(5);(6);()ab,,(2)a,，，322(7);(8);(5),,b()xy,3423(9);(10);(2),,x(3)ab,2、判斷下列式子是否正確，若不對(duì)，請(qǐng)糾正:mm22mm22，(1);(2);()aa,()aa,mnmn，mnmnaaa，,aaa,(3);(4).課后鞏固提高1、下列計(jì)算正確的是()336435510235xxx，,xxx,,A.B.C.D.()xx,xyxy,()2、81×27可以記為()367129333A.B.C.D.5a、3可以等于()2342332A.B.C.D.()(),,,aa()(),,,aa(),,aa()(),,,aa2324、計(jì)算的結(jié)果是(),,,bb()811811,b,bbbA.B.C.D.23105、在等式中，括號(hào)內(nèi)的代數(shù)式應(yīng)當(dāng)是()aaa,,,()104567A.B.C.D.aaaa221nn，6、若是正整數(shù)，當(dāng)時(shí)，等于()a,,1,,()anA.1B.,1C.0D.1或,12113nn,，7、計(jì)算的結(jié)果為()()xxx,,3n，36n，312n6n，6xxxxA.B.C.D.2436323，，8、，，.()a,a,(),,(2)ab，，mnmn，xx，39、已知，則;已知，則x=.a,2a,3a,42,10、計(jì)算:4433222(1);(2);(3);aa,,,，，，，,x,xy,，，，，11、下列各式中，正確的是()4485525A.mmm,,B.mmm,,23396612C.mmm,,D.yy,,2y12、下列各式中錯(cuò)誤的是()236284,2aA.16aB.()=,,，，，，x,y,x,y3113,,363263abC.D.,,mn,,mn，，,ab,,,327,,nn,，11,,cc13、已知n是大于1的自然數(shù),則等于()，，，，22nn,12n,2nccA.B.C.D.,c，，,c44a,a14、下列運(yùn)算中與結(jié)果相同的是()4444282224aaaaaA.B.C.D.，，a，，，，，，15、用簡(jiǎn)便方法計(jì)算1111279,,,,911，，()，1.5(1)(2)，，，,11(1),,,,,3916,,,,mm163927，，,316、已知,求m的值.11229nnx，,4217、若，解關(guān)于的方程.x216(2)，,mnm，2n18、若，，求的值(2,52,62?2指數(shù)擴(kuò)充及其運(yùn)算性質(zhì)1、將b寫(xiě)成分?jǐn)?shù)指數(shù)冪的形式:3,2mn2b,5b,4b,4(1);(2);(3).2、將分?jǐn)?shù)指數(shù)冪寫(xiě)成根式的形式:(1);(2);(3);(4).3、將根式寫(xiě)成分?jǐn)?shù)指數(shù)冪的形式:13322234()ab,xmn，(1);(2);(3);(4).3x4、計(jì)算:122311,,3224(1);(2);(3);(4).10098164,,,,,，,,,,2,55、已知，，求，，，.10103,101010104,11,,122,22aa，,36、已知，求，.aa，aa，?3指數(shù)函數(shù)xx1,n,m,01、已知，則指數(shù)函數(shù)1.，2.的圖像為()y,my,nxxxx2、如圖是指數(shù)函數(shù)的圖像則的關(guān)系是()a,b,c,dy,a,y,b,y,c,y,da,b,1,c,db,a,1,d,cA.B.1,a,b,c,da,b,1,d,cC.D.aba2,2,3a,b,03、已知，則的大小關(guān)系是()第2題ababaabaaaab223,,232,,223,,232,,A(B(C(D(a,aa,1,a,0,S,2,P,2,Q,0.24、若，則下列選項(xiàng)成立的是()13A.B.C.D.S,P,QP,Q,SQ,P,SS,Q,P,1.51,,5、設(shè)，則()0.90.48,,,yyy4,8,,,1232,,A.B.C.D.yyy,,yyy,,yyy,,yyy,,31221313212322xx6、若，那么的值為()x，13，9,10,3A.1B.2C.5D.1或50.90.80.87、已知?jiǎng)t的大小關(guān)系為.abc,,a,0.8,b,0.9,c,0.82xx8、解方程.32330,,,,?4.1對(duì)數(shù)及其運(yùn)算1、把下列指數(shù)式寫(xiě)成對(duì)數(shù)式:1,13,1)(2)(28,22113m,27,()5.73(3)(4)332、把下列對(duì)數(shù)式寫(xiě)成指數(shù)式:(1)log92,(2)log1253,5311log2,,log4,,(3)(4)234813、求下列各式中x的值:2x,,loglog86,(1)(2)64x32lnex,(3)(4)lg100,x4、求下列各式的值:1loglog125(1)(2)2516lg1000lg0.001(3)(4)14(5)(6)log15log1150.4log4lg1052210(7)(8)5、基礎(chǔ)練習(xí)(1)(2)log18log2,,lg2lg5，,336、加強(qiáng)鞏固lg2lg5lg8，,3(1)(2)12loglog20log4og，，,151515152lg50lg40,7lg4lg51，,(3)1142lglg7lg18g,，,(4)32lg0.5lg8，log2lg32,1010log1,，,(5)(6)lg2lg2lg5lg5，,，52xy22abc,,loglog()xylog(9)xyax,logby,logcz,log7、已知，，，請(qǐng)分別用表示式子，，.2222223z15?4.2換底公式1、求下列各式的值:log641log27(1)(2)(3)log2719163lg243(4)(5)(6)log9log32,log16log81,89932lg92、加強(qiáng)鞏固log134(1)log2log274，，(2)(log3log3)(log2log2)，，4839293、綜合應(yīng)用6blg3,blg2,a(1)設(shè),，試用、表示，，.alog4log12lg5321621xy(2)已知求.3436,,，xy?5對(duì)數(shù)函數(shù)1、求下列函數(shù)的定義域:1(1);(2);yx,，log(1)y,log22x,1yx,logyx,，log1(3);(4).232、求下列函數(shù)的反函數(shù):yx,log(1);(2);yx,log122x(3);(4)yx,2;y,32x(5);(6).yx,,log(1)y,523、比較各題中各數(shù)的大小:(1)，;(2)，log0.1;log3log5log2220.20.2(3)，;(4)log2，.log3log2log3(0,1)aa,,23aa17x,22,1,,x4、已知函數(shù)，則.fx(),ff(2),,,,,,log,1xx,2x,1,22,1,,xfx(),5、已知函數(shù)，且，則.fa()3,,fa(6),,,,，,log(1),1xx2,第三章指數(shù)函數(shù)和對(duì)數(shù)函數(shù)單元測(cè)試卷滿分150分，考試時(shí)間120分鐘一、選擇題(每小題6分，共60分()1(已知x，y為正實(shí)數(shù)，則()lglglglgxyxy，lg()lglgxyxy，A(B(222,，222,,lglglglgxyxylg()lglgxyxyC(D(222,，222,,x2(若函數(shù)y,f(x)是函數(shù)y,a(a>0，a?1)的反函數(shù)且f(2),1，則f(x),()1x,2A(B(C(D(2logxlogx12x223(已知，則函數(shù)y,f(x，1)在區(qū)間[2,8]上的最大