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1、2.6對(duì)數(shù)與對(duì)數(shù)函數(shù)知識(shí)梳理考點(diǎn)自測(cè)1.對(duì)數(shù)的概念(1)根據(jù)下圖的提示填寫與對(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)算法則如果a0,且a1,M0,N0,那么loga(MN)=;指數(shù)對(duì)數(shù)冪真數(shù)底數(shù)a0,且a1logaM+logaN logaM-logaN 知識(shí)梳理考點(diǎn)自測(cè)知識(shí)梳理考點(diǎn)自測(cè)4.對(duì)數(shù)函數(shù)的圖象與性質(zhì)(0,+)(1,0)增函數(shù)減函數(shù)知識(shí)梳理考點(diǎn)自測(cè)5.反函數(shù)指數(shù)函數(shù)y=ax(a0,且a1)與對(duì)數(shù)函數(shù)(a0,且a1)互為反函數(shù),它們的圖象關(guān)于直線對(duì)稱.y=logax y=x 知識(shí)梳理考點(diǎn)自測(cè)1.對(duì)數(shù)的性質(zhì)(a0,且a1,M0,b0)(1)loga1=0;
2、(2)logaa=1;(3)logaMn=nlogaM(nR);2.換底公式的推論(1)logablogba=1,即logab=(2)logablogbclogcd=logad.3.對(duì)數(shù)函數(shù)的圖象與底數(shù)大小的比較如圖,直線y=1與四個(gè)函數(shù)圖象交點(diǎn)的橫坐標(biāo)即為相應(yīng)的底數(shù).知識(shí)梳理考點(diǎn)自測(cè)知識(shí)梳理考點(diǎn)自測(cè) 知識(shí)梳理考點(diǎn)自測(cè)A.abcB.acbC.cabD.cb0,且a1)的值域?yàn)閥|00,且a1)的值域?yàn)閥|0y1,則0a0,且a1)的圖象恒過點(diǎn).(3,1)解析解析:當(dāng)4-x=1,即x=3時(shí),y=loga1+1=1.所以函數(shù)的圖象恒過點(diǎn)(3,1).考點(diǎn)一考點(diǎn)二考點(diǎn)三對(duì)數(shù)式的化簡(jiǎn)與求值對(duì)數(shù)式的化簡(jiǎn)與
3、求值例1化簡(jiǎn)下列各式:思考對(duì)數(shù)運(yùn)算的一般思路是什么?考點(diǎn)一考點(diǎn)二考點(diǎn)三解題心得對(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.考點(diǎn)一考點(diǎn)二考點(diǎn)三D4考點(diǎn)一考點(diǎn)二考點(diǎn)三對(duì)數(shù)函數(shù)的圖象及其應(yīng)用對(duì)數(shù)函數(shù)的圖象及其應(yīng)用 CB考點(diǎn)一考點(diǎn)二考點(diǎn)三考點(diǎn)一考點(diǎn)二考點(diǎn)三思考應(yīng)用對(duì)數(shù)型函數(shù)的圖象主要解決哪些問題?解題心得應(yīng)用對(duì)數(shù)型函數(shù)的圖象可求解的問題:(1)對(duì)一些可通過平移、對(duì)稱變換作出其圖象的對(duì)數(shù)型函數(shù),在求解其單調(diào)性
4、(單調(diào)區(qū)間)、值域(最值)、零點(diǎn)時(shí),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.考點(diǎn)一考點(diǎn)二考點(diǎn)三對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練2(1)(2017福建泉州一模,文7)函數(shù)f(x)=ln(x+1)+ln(x-1)+cosx的圖象大致是()AD考點(diǎn)一考點(diǎn)二考點(diǎn)三解析解析:(1)函數(shù)f(x)=ln(x+1)+ln(x-1)+cosx,則函數(shù)的定義域?yàn)閤1,故排除C,D;-1cosx1,當(dāng)x+時(shí),f(x)+,故選A.設(shè)曲線y=x2-2x在x=0處的切線l的斜率為k,由y=2x-2,可知k=y|x=0=-2.要使|f(x)|ax,則直線y=ax的傾斜角要大于等于
5、直線l的傾斜角,小于等于,即a的取值范圍是-2,0.考點(diǎn)一考點(diǎn)二考點(diǎn)三對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用(多考向多考向)考向1比較含對(duì)數(shù)的函數(shù)值的大小例3(2017天津,文6)已知奇函數(shù)f(x)在R上是增函數(shù),若a=-f ,b=f(log24.1),c=f(20.8),則a,b,c的大小關(guān)系為()A.abcB.bacC.cbaD.calog24.1log24=2,20.8log24.120.8.又f(x)在R上是增函數(shù),f(log25)f(log24.1)f(20.8),即abc.故選C.思考如何比較兩個(gè)含對(duì)數(shù)的函數(shù)值的大小?考點(diǎn)一考點(diǎn)二考點(diǎn)三考向2解含對(duì)數(shù)的函數(shù)不等式 CC考點(diǎn)一考
6、點(diǎn)二考點(diǎn)三思考如何解簡(jiǎn)單對(duì)數(shù)不等式?考點(diǎn)一考點(diǎn)二考點(diǎn)三考向3對(duì)數(shù)型函數(shù)的綜合問題例5已知f(x)=loga(ax-1)(a0,且a1).(1)求f(x)的定義域;(2)討論函數(shù)f(x)的單調(diào)性.解(1)由ax-10,得ax1.當(dāng)a1時(shí),x0;當(dāng)0a1時(shí),x1時(shí),f(x)的定義域?yàn)?0,+);當(dāng)0a1時(shí),設(shè)0 x1x2,考點(diǎn)一考點(diǎn)二考點(diǎn)三思考在判斷對(duì)數(shù)型復(fù)合函數(shù)的單調(diào)性時(shí)需要注意哪些條件?解題心得1.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)的圖象;底數(shù)、真數(shù)均不同的可借助中間值(0或1).2.
7、解簡(jiǎn)單對(duì)數(shù)不等式,先統(tǒng)一底數(shù),再利用函數(shù)的單調(diào)性,要注意對(duì)底數(shù)a的分類討論.3.在判斷對(duì)數(shù)型復(fù)合函數(shù)的單調(diào)性時(shí),一定要明確底數(shù)a對(duì)增減性的影響,以及真數(shù)必須為正的限制條件.考點(diǎn)一考點(diǎn)二考點(diǎn)三對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練3(1)已知定義在R上的函數(shù)f(x)=2|x-m|-1(m為實(shí)數(shù))為偶函數(shù).記a=f(log0.53),b=f(log25),c=f(2m),則a,b,c的大小關(guān)系為()A.abcB.acbC.cabD.cba(2)(2017河北武邑中學(xué)一模,文7)已知f(x)=是奇函數(shù),則使f(x)0,且a1.求f(x)的定義域;判斷f(x)的奇偶性,并予以證明;當(dāng)a1時(shí),求使f(x)0的x的取值范圍.CA考點(diǎn)一考點(diǎn)二考點(diǎn)三考點(diǎn)一考點(diǎn)二考點(diǎn)三考點(diǎn)一考點(diǎn)二考點(diǎn)三1.多個(gè)對(duì)數(shù)函數(shù)圖象比較底數(shù)大小的問題,可通過圖象與直線y=1交點(diǎn)的橫坐標(biāo)進(jìn)行判定.2.研究對(duì)數(shù)型函數(shù)的圖象時(shí),一般從最基本的對(duì)數(shù)函數(shù)的圖象入手,通過平移、伸縮、對(duì)稱變換得到.特別地,要注意底數(shù)a1和0a1的兩種不同情況.有些復(fù)雜的問題,借助于函數(shù)圖象來(lái)解決,就變得簡(jiǎn)單了,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，然后根據(jù)單調(diào)性來(lái)解決.考點(diǎn)一考點(diǎn)二考點(diǎn)三
（福建專）高考數(shù)學(xué)一輪復(fù)習(xí) 2.6 對(duì)數(shù)與對(duì)數(shù)函數(shù)課件 文
