1、2.5對(duì)數(shù)與對(duì)數(shù)函數(shù)核心考點(diǎn)精準(zhǔn)研析考點(diǎn)一對(duì)數(shù)式的化簡(jiǎn)與求值1.(2019北京高考)在天文學(xué)中,天體的明暗程度可以用星等或亮度來(lái)描述.兩顆星的星等與亮度滿足m2-m1=lg,其中星等為mk的星的亮度為Ek(k=1,2).已知太陽(yáng)的星等是-26.7,天狼星的星等是-1.45,則太陽(yáng)與天狼星的亮度的比值為()A.1010.1B.10.1C.lg10.1D.10-10.12.(2020深圳模擬)設(shè)函數(shù)y=f(x)的圖象與y=2x+a的圖象關(guān)于直線y=-x對(duì)稱,且f(-2)+f(-4)=1,則a=()A.-1B.1C.2D.43.設(shè)x,y,z為正數(shù),且2x=3y=5z,則()A.2x3y5zC.3y5

2、z2xB.5z2x3yD.3y2xlogm215logm563y2x0,且a1)是解決有關(guān)指數(shù)、對(duì)數(shù)問(wèn)題的有效方法,在運(yùn)算中應(yīng)注意互化.(4)利用換底公式將不同底的對(duì)數(shù)式轉(zhuǎn)化為同底的對(duì)數(shù)式.考點(diǎn)二對(duì)數(shù)函數(shù)的圖象及其應(yīng)用【典例】1.已知函數(shù)y=loga(x+c)(a,c為常數(shù),其中a0,且a1)的圖象如圖,則下列結(jié)論成立的是()A.a1,c1B.a1,0c1C.0a1D.0a1,0c0,且a1)的圖象可能是()3.已知函數(shù)f(x)=_.【解題導(dǎo)思】g(x)=log2x,則f(x)與g(x)兩函數(shù)圖象的交點(diǎn)個(gè)數(shù)為序號(hào)123聯(lián)想解題由圖象是下降的,想到對(duì)數(shù)的底數(shù)0a1由y=與y=loga,想到指數(shù)函

3、數(shù)與對(duì)數(shù)函數(shù)的圖象由兩函數(shù)圖象的交點(diǎn)個(gè)數(shù),想到畫出兩個(gè)函數(shù)的圖象【解析】1.選D.由題圖可知,函數(shù)在定義域內(nèi)為減函數(shù),所以0a0,即logac0,所以0c1.2.選D.當(dāng)0a1時(shí),函數(shù)y=ax的圖象過(guò)定點(diǎn)(0,1)且單調(diào)遞增,則函數(shù)y=的圖象過(guò)定點(diǎn)(0,1)且單調(diào)遞減,函數(shù)y=loga的圖象過(guò)定點(diǎn)且單調(diào)遞增,各選項(xiàng)均不符合.3.如圖,函數(shù)g(x)的圖象與函數(shù)f(x)的圖象交于兩點(diǎn),且均在函數(shù)y=8x-8(x1)的圖象上.答案:21.應(yīng)用對(duì)數(shù)型函數(shù)的圖象可求解的問(wèn)題(1)對(duì)一些可通過(guò)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(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)

4、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)圖象問(wèn)題,利用數(shù)形結(jié)合法求解.2.對(duì)數(shù)函數(shù)圖象的規(guī)律在第一象限內(nèi),不同底的對(duì)數(shù)函數(shù)的圖象從左到右底數(shù)逐漸增大.1.已知函數(shù)f(x)=loga(2x+b-1)(a0,a1)的圖象如圖所示,則a,b滿足的關(guān)系是()A.0a-1b1B.0ba-11C.0b-1a1D.0a-1b-11.函數(shù)圖象與y軸的交點(diǎn)坐標(biāo)為(0,logab),由函數(shù)圖象可知-1logab0,即logaa-1logabloga1,所以a-1b1.綜上有0a-1b1.2.(2020北京模擬)已知函數(shù)f(x)=2x(x0)與g(x)=ln(x+a)的圖象上存在關(guān)于y軸對(duì)稱的點(diǎn),則a的取值范圍

5、是()A.(-,2)C.(2,e)B.(-,e)D.(e,+)【解析】選B.在同一直角坐標(biāo)系中作出函數(shù)f(x)=2x(x0)個(gè)單位長(zhǎng)度,恰好過(guò)(0,1)時(shí),函數(shù)f(x)與g(x)就不存在關(guān)于y軸對(duì)稱的點(diǎn),所以0ae,當(dāng)y=lnx向右平移|a|(a0)個(gè)單位長(zhǎng)度,函數(shù)f(x)與g(x)總存在關(guān)于y軸對(duì)稱的點(diǎn),當(dāng)a=0時(shí),顯然滿足題意,綜上:ae.命題精解讀學(xué)霸好方法考點(diǎn)三對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用考什么:(1)求對(duì)數(shù)函數(shù)的單調(diào)性,利用對(duì)數(shù)函數(shù)的單調(diào)性比較大小、求值或解不等式、求參數(shù)值等問(wèn)題.(2)考查數(shù)學(xué)運(yùn)算、直觀想象、邏輯推理等核心素養(yǎng).怎么考:對(duì)數(shù)函數(shù)奇偶性、單調(diào)性,函數(shù)的周期性以及對(duì)稱性等知識(shí)

6、單獨(dú)或交匯考查,也可能以分段函數(shù)的形式呈現(xiàn).新趨勢(shì):對(duì)數(shù)函數(shù)的圖象與對(duì)稱性、交點(diǎn)個(gè)數(shù)、不等式交匯考查.1.比較對(duì)數(shù)式的大小的方法(1)能化成同底數(shù)的先化成同底對(duì)數(shù)值,再利用單調(diào)性比較大小.(2)不能化成同底數(shù)的,一般引入“1”“0”“-1”等中間量比較大小.(3)在研究對(duì)數(shù)型函數(shù)的單調(diào)性時(shí),當(dāng)?shù)讛?shù)a與“1”的大小關(guān)系不確定時(shí),要分類討論.2.對(duì)數(shù)函數(shù)單調(diào)性的判斷(1)求單調(diào)區(qū)間必須先求定義域.(2)根據(jù)對(duì)數(shù)的底數(shù)a進(jìn)行判斷,0a1時(shí)為增函數(shù).(3)對(duì)數(shù)型函數(shù)的單調(diào)性根據(jù)復(fù)合函數(shù)“同增異減”進(jìn)行判斷.比較大小問(wèn)題【典例】(2019全國(guó)卷)已知a=log20.2,b=20.2,c=0.20.3,則

7、()A.abcC.cabB.acbD.bca【解析】選B.a=log20.220=1,00.20.30.20=1,則0c1,所以acb.如何比較指數(shù)式與對(duì)數(shù)式的大小?提示:數(shù)形結(jié)合或找中間量(如1,0,-1等),再結(jié)合函數(shù)單調(diào)性比較大小.與對(duì)數(shù)函數(shù)有關(guān)的不等式問(wèn)題【典例】當(dāng)0 x時(shí),4xlogax,則a的取值范圍是()A.B.C.(1,)D.(,2)【解析】選B.由題意知0a2,解得a,所以a1.一邊為指數(shù)式,另一邊為對(duì)數(shù)的不等式如何求解?提示:將兩邊分別看成一個(gè)函數(shù),畫出兩個(gè)函數(shù)的圖象,結(jié)合圖象的交點(diǎn)求解.對(duì)數(shù)函數(shù)性質(zhì)的綜合應(yīng)用【典例】已知函數(shù)f(x)=lnx+ln(2-x),則()A.f(

8、x)在(0,2)上單調(diào)遞增B.f(x)在(0,2)上單調(diào)遞減C.f(x)的圖象關(guān)于直線x=1對(duì)稱D.f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱【解析】選C.由題意知,f(2-x)=ln(2-x)+lnx=f(x),所以f(x)的圖象關(guān)于直線x=1對(duì)稱,C正確,D錯(cuò)誤;又f(x)=-=(0 x2),在(0,1)上單調(diào)遞增,在(1,2)上單調(diào)遞減,A,B錯(cuò)誤.【一題多解】解決本題還可以采用以下方法:選C.由題意知,f(x)=lnx+ln(2-x)的定義域?yàn)?0,2),f(x)=+=,由得0 x1;由得1x2,所以函數(shù)f(x)=lnx+ln(2-x)在(0,1)上單調(diào)遞增,在(1,2)上單調(diào)遞減,所以排除A

9、,B;又f=ln+ln=ln,f=ln+ln=ln,所以f=f=ln,所以排除D.如何求解對(duì)數(shù)函數(shù)性質(zhì)的綜合問(wèn)題?提示:認(rèn)真聯(lián)想對(duì)數(shù)函數(shù)的各個(gè)性質(zhì)的定義及其作用,在其交匯點(diǎn)處尋找突破口.1.已知函數(shù)f(x)=loga|x|在(0,+)上單調(diào)遞增,則f(-2)_f(a+1).(填“”)【解析】因?yàn)閒(x)=loga|x|在(0,+)上單調(diào)遞增,所以a1,所以a+12.因?yàn)閒(x)是偶函數(shù),所以f(-2)=f(2)f(a+1).答案:2.(2019濰坊模擬)已知函數(shù)f(x)=若f(2-a)=1,則f(a)=_.【解析】當(dāng)2-a0時(shí),f(2-a)=-log2(1+a)=1.解得a=-,不合題意.當(dāng)2

10、-a2,即a0時(shí),f(2-a)=2-a-1=1,即2-a=2,解得a=-1,所以f(a)=f(-1)=-log24=-2.答案:-21.(2019綿陽(yáng)模擬)若x,y,zR+,且3x=4y=12z,(n,n+1),nN,則n的值是(A.2B.3C.4D.5【解析】選C.設(shè)3x=4y=12z=t(t1),則x=log3t,y=log4t,z=log12t,所以=+=log312+log412=2+log34+log43.因?yàn)?log342,0log431,所以1log34+log432=2,所以42+log34+log435,即(4,5).所以n=4.2.(2020揚(yáng)州模擬)設(shè)f(x)=a=0.7-0.5,b=log0.50.7,c=log0.75,則f(a),f(b),f(c)的大小關(guān)系為_(kāi).【解析】當(dāng)x0時(shí),f(x)=x+1是單調(diào)增函數(shù),所以有