2025版高考數(shù)學(xué)一輪總復(fù)習(xí)10年高考真題分類題組2.4對(duì)數(shù)與對(duì)數(shù)函數(shù)

.4對(duì)數(shù)與對(duì)數(shù)函數(shù)考點(diǎn)一對(duì)數(shù)的概念及運(yùn)算1.(2024課標(biāo)Ⅲ文,10,5分)設(shè)a=log32,b=log53,c=23,則()A.a<c<bB.a<b<cC.b<c<aD.c<a<b答案A因?yàn)閍=log32=log338<log339=b=log53=log5327>log5325=23=c,所以2.(2024課標(biāo)Ⅲ理,12,5分)設(shè)a=log0.20.3,b=log20.3,則()A.a+b<ab<0B.ab<a+b<0C.a+b<0<abD.ab<0<a+b答案B本題考查不等式及對(duì)數(shù)運(yùn)算.解法一:∵a=log0.20.3>log0.21=0,b=log20.3<log21=0,∴ab<0,解除C.∵0<log0.20.3<log0.20.2=1,log20.3<log20.5=-1,即0<a<1,b<-1,∴a+b<0,解除D.∵ba=log20.3log0.20.3=lg0.2lg2=log20.2,∴b-解法二:易知0<a<1,b<-1,∴ab<0,a+b<0,∵1a+1b=log0.30.2+log0.32=log即a+∴ab<a+b<0.故選B.方法總結(jié)比較代數(shù)式大小的常用方法(1)作差法:其基本步驟為作差、變形、推斷符號(hào)、得出結(jié)論.用作差法比較大小的關(guān)鍵是推斷差的正負(fù).變形常采納配方、因式分解、分子(分母)有理化等方法.(2)作商法:即通過推斷商與1的大小關(guān)系,得出結(jié)論.要特殊留意當(dāng)商與1的大小確定后,必需對(duì)商式分子、分母的正負(fù)進(jìn)行推斷,這是用作商法比較大小時(shí)最簡單漏掉的關(guān)鍵步驟.(3)單調(diào)性法:利用有關(guān)函數(shù)的單調(diào)性比較大小.(4)特值驗(yàn)證法:對(duì)于一些給出取值范圍的題目,可采納特值驗(yàn)證法比較大小.3.(2024天津理,5,5分)已知a=log2e,b=ln2,c=log1213,則a,b,cA.a>b>cB.b>a>cC.c>b>aD.c>a>b答案D本題主要考查對(duì)數(shù)的大小比較.由已知得c=log23,∵log23>log2e>1,b=ln2<1,∴c>a>b,故選D.考點(diǎn)二對(duì)數(shù)函數(shù)的圖象與性質(zhì)1.(2024天津文,5,5分)已知a=log27,b=log38,c=0.30.2,則a,b,c的大小關(guān)系為()A.c<b<aB.a<b<cC.b<c<aD.c<a<b答案A本題考查指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的圖象和性質(zhì);通過對(duì)對(duì)數(shù)式的估算或適當(dāng)“縮放”考查學(xué)生的直觀想象與邏輯推理的核心素養(yǎng).明顯c=0.30.2∈(0,1).因?yàn)閘og33<log38<log39,所以1<b<2.因?yàn)閘og27>log24=2,所以a>2.故c<b<a.選A.2.(2017課標(biāo)Ⅱ文,8,5分)函數(shù)f(x)=ln(x2-2x-8)的單調(diào)遞增區(qū)間是()A.(-∞,-2)B.(-∞,1)C.(1,+∞)D.(4,+∞)答案D本題主要考查復(fù)合函數(shù)的單調(diào)性.由x2-2x-8>0可得x>4或x<-2,所以x∈(-∞,-2)∪(4,+∞),令u=x2-2x-8,則其在x∈(-∞,-2)上單調(diào)遞減,在x∈(4,+∞)上單調(diào)遞增.又因?yàn)閥=lnu在u∈(0,+∞)上單調(diào)遞增,所以y=ln(x2-2x-8)在x∈(4,+∞)上單調(diào)遞增.故選D.易錯(cuò)警示本題易忽視定義域而錯(cuò)選C.方法總結(jié)復(fù)合函數(shù)的單調(diào)性符合同增異減的原則.3.(2016課標(biāo)Ⅰ文,8,5分)若a>b>0,0<c<1,則()A.logac<logbcB.logca<logcbC.ac<bcD.ca>cb答案B∵0<c<1,∴當(dāng)a>b>1時(shí),logac>logbc,A項(xiàng)錯(cuò)誤;∵0<c<1,∴y=logcx在(0,+∞)上單調(diào)遞減,又a>b>0,∴l(xiāng)ogca<logcb,B項(xiàng)正確;∵0<c<1,∴函數(shù)y=xc在(0,+∞)上單調(diào)遞增,又∵a>b>0,∴ac>bc,C項(xiàng)錯(cuò)誤;∵0<c<1,∴y=cx在(0,+∞)上單調(diào)遞減,又∵a>b>0,∴ca<cb,D項(xiàng)錯(cuò)誤.故選B.評(píng)析本題主要考查了冪函數(shù)、指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的圖象和性質(zhì),嫻熟駕馭冪函數(shù)、指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的圖象和性質(zhì)是解題的關(guān)鍵.4.(2015課標(biāo)Ⅱ理,5,5分)設(shè)函數(shù)f(x)=1+lo則f(-2)+f(log212)=()A.3B.6C.9D.12答案C∵-2<1,∴f(-2)=1+log2[2-(-2)]=3;∵log212>1,∴f(log212)=2log212-5.(2013課標(biāo)Ⅱ理,8,5分)設(shè)a=log36,b=log510,c=log714,則()A.c>b>aB.b>c>aC.a>c>bD.a>b>c答案D由對(duì)數(shù)運(yùn)算法則得a=log36=1+log32,b=1+log52,c=1+log72,由對(duì)數(shù)函數(shù)圖象得log32>log52>log72,所以a>b>c,故選D.6.(2013課標(biāo)Ⅱ文,8,5分)設(shè)a=log32,b=log52,c=log23,則()A.a>c>bB.b>c>aC.c>b>aD.c>a>b答案D∵3<2<3,1<2<5,3>2,∴l(xiāng)og33<log32<log33,log51<log52<log55,log23>log22,∴12<a<1,0<b<1∴c>a>b.故選D.7.(2012課標(biāo)文,11,5分)當(dāng)0<x≤12時(shí),4x<logax,則a的取值范圍是(A.0,22B.22,1C.(1,答案B易知0<a<1,則函數(shù)y=4x與y=logax的大致圖象如圖,則只需滿意loga12>2,解得a>22,∴22評(píng)析本題考查了利用數(shù)形結(jié)合解指數(shù)、對(duì)數(shù)不等式.8.(2024課標(biāo)Ⅰ