課時(shí)規(guī)范練10對數(shù)與對數(shù)函數(shù)基礎(chǔ)鞏固組1.(2022湖南長郡中學(xué)一模)已知集合A={x|y=ln(x-2)},集合B=yy=(12)x,A.? B.(2,8)C.(3,8) D.(8,+∞)2.(2022浙江,7)已知2a=5,log83=b,則4a-3b=()A.25 B.5 C.259 D.3.若函數(shù)y=a|x|(a>0,且a≠1)的值域?yàn)閧y|y≥1},則函數(shù)y=loga|x|的圖象大致是()4.已知a=log37,b=log25343,c=12+4log92,則()A.b>a>c B.c>a>bC.a>b>c D.b>c>a5.已知log23=a,3b=7,則log2156=()A.ab+3a+C.ab+3a+6.已知3x=2y=t,且1x+1y=2,則A.26 B.6 C.36 D.67.若函數(shù)f(x)=lg(ax2-2x+a)的值域?yàn)镽,則實(shí)數(shù)a的取值范圍為()A.(-1,0) B.(0,1)C.[0,1] D.(1,+∞)8.log23·log34-(3)log9.若函數(shù)f(x)=logax(a>1)在區(qū)間[a,2a]上的最大值是最小值的3倍,則a=.

10.方程(log3x)2+log93x=2的解集為.

綜合提升組11.已知a=log0.20.3,b=log0.30.2,c=log23,則a,b,c的大小關(guān)系為()A.a<b<c B.b<c<aC.c<a<b D.a<c<b12.已知函數(shù)f(x)=log2(-x2-mx+16)在[-2,2]上單調(diào)遞減,則m的取值范圍是()A.[4,+∞) B.(-6,6)C.(-6,4] D.[4,6)13.若函數(shù)y=f(x)與y=3-x的圖象關(guān)于直線y=x對稱,則函數(shù)y=f(4x-x2)的單調(diào)遞增區(qū)間為()A.(2,4) B.(0,2)C.(-∞,2) D.(2,+∞)14.已知f(x)=2+log3x,x∈[1,9],則y=[f(x)]2+f(x2)的值域?yàn)?)A.[6,23] B.[6,13]C.[4,11] D.[4,20]創(chuàng)新應(yīng)用組15.(2022山西太原一模)已知實(shí)數(shù)x,y滿足x·2x=7,y(log2y-2)=28,則xy=()A.112 B.28 C.7 D.416.(2022全國甲,文12)已知9m=10,a=10m-11,b=8m-9,則()A.a>0>b B.a>b>0C.b>a>0 D.b>0>a

參考答案課時(shí)規(guī)范練10對數(shù)與對數(shù)函數(shù)1.B∵A={x|y=ln(x-2)}={x|x>2},B=yy=(12)

x,x>-3={2.C由log83=b,得8b=3,即23b=3,則2a-3b=2a23b=53,所以4a-33.B若函數(shù)y=a|x|(a>0,且a≠1)的值域?yàn)閧y|y≥1},則a>1,且函數(shù)y=loga|x|是偶函數(shù),故它的圖象大致如選項(xiàng)B中圖示.4.Ab=log25343=3log37log325>log37=a,c=12+4log92=log948<log9495.A由3b=7,可得log37=b,所以log2156=log6.B根據(jù)題意,3x=2y=t>0,則有x=log3t,y=log2t,則1x=logt3,1y=logt2.又1x+1y=2,即logt3+logt2=logt6=2,所以t2=6,解得t=±6,因?yàn)?.C由題意,函數(shù)f(x)=lg(ax2-2x+a)的值域?yàn)镽,根據(jù)對數(shù)函數(shù)的性質(zhì),可得轉(zhuǎn)化為g(x)=ax2-2x+a的值域能取到(0,+∞)內(nèi)的任意實(shí)數(shù),當(dāng)a=0,則g(x)=-2x,函數(shù)g(x)的值域?yàn)镽,滿足題意;當(dāng)a≠0,要使得g(x)的值域能取到(0,+∞)上的任意實(shí)數(shù),則滿足a>0,Δ=(-2)2-4a8.2-2log23·log34-(3)log32=log23·2log239.2∵a>1,∴函數(shù)f(x)在區(qū)間[a,2a]上為增函數(shù),由已知條件可得loga(2a)=3logaa=logaa3,∴a3=2a,∵a>1,解得a=2.10.3,39∵(log3x)2+log93x=2,∴(log3x)2+12log33x=(log3x)2+12(log33+log3x)=2,即(log3x)2+12log3x-32=0,令t=log3x,則方程可化為t2+12t-32=0,解得t=1或t=-32,∴x=3或x=3-32,即x=3或x=39.∴方程11.A因?yàn)?<a=log0.20.3<1,b=log0.30.2>1,c=log23>1,又bc=log0.30.2·log32=lg2-1lg3-1·lg2lg3=lg22-lg2lg23-lg3,因?yàn)楹瘮?shù)f(x)=x2-x=x-122?14在0,12上單調(diào)遞減,且f(0)=0,又因?yàn)?2>lg12.D令g(x)=-x2-mx+16,因?yàn)閥=log2x是增函數(shù),所以要使f(x)在[-2,2]上單調(diào)遞減,只需g(x)在[-2,2]上單調(diào)遞減,且g(x)>0恒成立.故g(x)min=13.A函數(shù)y=f(x)與y=3-x的圖象關(guān)于直線y=x對稱,可知它們互為反函數(shù),∴y=f(x)=log13x,則f(4x-x2)=log13(4x-x2),令t=4x-x2,∵t>0,∴0<x<4.∵f(x)在其定義域內(nèi)是減函數(shù),而t=4x-x2在(0,2)上單調(diào)遞增,在(2,4)上單調(diào)遞減,則復(fù)合函數(shù)y=f(4x-x214.B因?yàn)閒(x)=2+log3x,x∈[1,9],所以y=[f(x)]2+f(x2)的定義域?yàn)?≤x≤9,1≤x2≤9,解得1≤x≤3,所以該函數(shù)的定義域?yàn)閇1,3].所以0≤log3x≤1.所以y=[f(x)]2+f(x2)=(2+log3x)2+(2+log3x2)=(log3x+3)2-3,t=log3x(0≤t≤1),所以y=(t+3)2-3(0≤t≤1),當(dāng)t=0時(shí),y=6,當(dāng)t=1時(shí),y=13,所以6≤15.B由y(log2y-2)=28,得y4log2y4=7,則2log2y4·log2y4=7,則log2y4>0,令f(x)=x·2x,x>0,則f'(x)=2x+x·2xln2=2x(1+xln2)>0,即f(x)在(0,+∞)上單調(diào)遞增.∵x·2x=log2y4·2log2y4=7,即f(x)=flog2y4=7,∴x=16.A由9m=10,得m=log910∈1,32.由于a=10m