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1、+2019年數(shù)學(xué)高考教學(xué)資料+第六節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)考點(diǎn)一對(duì)數(shù)式的化簡(jiǎn)與求值 例1(1)已知loga2m,loga3n,求a2mn;(2)計(jì)算;(3)計(jì)算(log32log92)(log43log83)自主解答(1)法一:loga2m,loga3n,am2,an3,a2mn(am)2an22312.法二:loga2m,loga3n,a2mn(am)2an(aloga2)2aloga322312.(2)原式1.(3)原式.【互動(dòng)探究】在本例(1)的條件下,求loga36的值解:loga36loga4loga922(mn)【方法規(guī)律】對(duì)數(shù)運(yùn)算的一般思路來(lái)源:(1)首先利用冪的運(yùn)算把底數(shù)或真數(shù)進(jìn)行變形
2、,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1計(jì)算100_.解析:原式20.答案:202設(shè)2a5bm,且2,則m_.解析:2a5bm,alog2m,blog5m,logm2logm5logm102.m210,m.答案:考點(diǎn)二對(duì)數(shù)函數(shù)的圖象及其應(yīng)用 例2(1)函數(shù)ylogax與yxa在同一坐標(biāo)系中的圖象可能是()ABCD (2)已知函數(shù)f(x)loga(2xb1)(a0,a1)的圖象如圖所示,則a,b滿足的關(guān)系是()A0a1b1B0ba11C0b1a1D0a1b11
3、.又由圖象知函數(shù)圖象與y軸交點(diǎn)的縱坐標(biāo)介于1和0之間,即1f(0)0,所以1logab0,故a1b1,因此0a1bf(a),則實(shí)數(shù)a的取值范圍是()A(1,0)(0,1)B(,1)(1,)C(1,0)(1,)D(,1)(0,1)來(lái)源:(4)(2014中山模擬)已知函數(shù)f(x)loga(8ax)(a0,a1),若f(x)1在區(qū)間1,2上恒成立,則實(shí)數(shù)a的取值范圍為_自主解答(1)要使有意義,需滿足x10且x10,得x1且x1.(2)23,12,32,log3log32log33,log51log52log5,log23log22,a1,0b,c1.cab.(3)由題意可得或解得a1或1a0.(4
4、)當(dāng)a1時(shí),f(x)loga(8ax)在1,2上是減函數(shù),由f(x)1恒成立,則f(x)minloga(82a)1,解得1a.若0a1時(shí),f(x)在x1,2上是增函數(shù),由f(x)1恒成立,則f(x)minloga(8a)1,且82a0,a4,且a4,故不存在綜上可知,實(shí)數(shù)a的取值范圍是.答案(1)C(2)D(3)C(4)對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用問(wèn)題的常見類型及解題策略(1)求函數(shù)的定義域要注意對(duì)數(shù)函數(shù)的底數(shù)和真數(shù)的取值范圍,列出對(duì)應(yīng)的不等式(組)求解即可(2)比較對(duì)數(shù)式的大小若底數(shù)為同一常數(shù),則可由對(duì)數(shù)函數(shù)的單調(diào)性直接進(jìn)行判斷;若底數(shù)為同一字母,則需對(duì)底數(shù)進(jìn)行分類討論若底數(shù)不同,真數(shù)相同,則可以
5、先用換底公式化為同底后,再進(jìn)行比較若底數(shù)與真數(shù)都不同,則常借助1,0等中間量進(jìn)行比較(3)解對(duì)數(shù)不等式形如logaxlogab的不等式,借助ylogax的單調(diào)性求解,如果a的取值不確定,需分a1與0a1兩種情況討論;形如logaxb的不等式,需先將b化為以a為底的對(duì)數(shù)式的形式1已知a5log23.4,b5log43.6,clog30.3,則()Aabc BbacCacb Dcab解析:選Ca5log23.4,b5log43.6,clog30.35log3.又log23.4log31,0log43.61,5log23.4log30.35log43.6,即acb.2(2014嘉興模擬)已知函數(shù)f(
6、x)loga(3ax)(1)當(dāng)x0,2時(shí),函數(shù)f(x)恒有意義,求實(shí)數(shù)a的取值范圍;(2)是否存在這樣的實(shí)數(shù)a,使得函數(shù)f(x)在區(qū)間1,2上為減函數(shù),并且最大值為1?如果存在,試求出a的值;如果不存在,請(qǐng)說(shuō)明理由解:(1)a0且a1,設(shè)t3ax,則t3ax為減函數(shù),x0,2時(shí),t最小值為32a.當(dāng)x0,2時(shí),f(x)恒有意義,即x0,2時(shí),3ax0恒成立32a0,即a0且a1,a(0,1).(2)t3ax,a0,函數(shù)t(x)在R上為減函數(shù)f(x)在區(qū)間1,2上為減函數(shù),ylogat為增函數(shù)a1,x1,2時(shí),t(x)最小值為32a,f(x)最大值為f(1)loga(3a),即故這樣的實(shí)數(shù)a不存在來(lái)源:課堂歸納通法領(lǐng)悟1種關(guān)系指數(shù)式與對(duì)數(shù)式的互化abNlogaNb(a0,a1,N0)2個(gè)注意點(diǎn)解決對(duì)數(shù)問(wèn)題應(yīng)注意的兩點(diǎn)解決與對(duì)數(shù)有關(guān)的問(wèn)題時(shí):(1)務(wù)必先研究函數(shù)的定義域;(2)對(duì)數(shù)函數(shù)的單調(diào)性取決于底數(shù)a,應(yīng)注意底數(shù)的取值范圍3個(gè)關(guān)鍵點(diǎn)對(duì)數(shù)函數(shù)圖象的畫法畫對(duì)數(shù)函數(shù)ylogax的圖象應(yīng)抓住三個(gè)關(guān)鍵點(diǎn):(a,1),(1,0),.4種方法對(duì)數(shù)值的大小比較方法(1) 化同底后利用函數(shù)的單調(diào)性;(2)作差或作商法;(3)利用中間量(0或1);(4)化同真數(shù)后利用圖象比較. 高考數(shù)學(xué)復(fù)習(xí)精品高考數(shù)學(xué)復(fù)習(xí)精品