高一数学下册过关检测试题4.doc

训练 19 对数

基础巩固 站起来,拿得到!

1.在 y=log(x-2)(5-x)中,实数 x 的取值范围是( A.x>5 或 x<2

) B.2<x<5

C.2<x<3 或 3<x<5

D.3<x<4

答案:C

?5 ? x ? 0 解析: ??x ? 2 ? 0 ? 2<x<3 或 3<x<5.
??x ? 2 ? 1

?1
2.已知 log7[log3(log2x)]=0,那么 x 2 等于( )

1

1

A.

B.

3

23

1
C.
22

1
D.
33

答案:C

解析:由外到里依次有

log3(log2x)=1,log2x=3,x=23,

x

?

1 2

?

?3
22

?

1

.

22

3.给出下列四个命题:①对数的真数是非负数;②若 a>0 且 a≠1,则 loga1=0;③若 a>0 且 a≠1,

logaa=1;④若 a>0,且 a≠1, aloga2 =2.其中正确的命题是( )

A.①②③

B.②③④

C.①③ 答案:B

D.①②③④

解析:①对数的真数是正数而不是非负数,其他几个是正确的.

4.(四川成都模拟)lg8+3lg5 的值为( )

A.-3

B.-1

C.1

D.3

答案:D

解析:原式=3lg2+3lg5=3lg10=3.

5.满足等式 2lg(3x-2)=lgx+lg(3x+2)的实数 x 的值为______________.

答案:2

?3x ? 2 ? 0

解析: ??x ? 0

?x> 2 ,(3x-2)2=x(3x+2).

??3x ? 2 ? 0

3

解得 x= 1 (舍)或 x=2. 3

6.















式:(1)lg(ab)=lga+lgb;(2)lg a =lga-lgb;(3) 1 lg( a )2=lg a ;(4)lg(ab)= 1 .

b

2b b

log ab 10

其中正确的命题的个数为_________________.

答案:0

解析:(1)(2)(3)(4)都是错误的,例如:

(1)lg[(-2)×(-3)]≠lg(-2)+lg(-3);

(2)lg ? 2 ≠lg(-2)-lg(-3); ?3

(3) 1 lg( 1 )2≠lg( 1 );

2 ?2

?2

(4)lg(2× 1 )≠ 1 .

2

log (2?1 ) 10

2

注意:在应用对数的性质时,一定要使运算过程中的每一个数式都有意义.

7.已知 log32=a,3b=5,试用 a、b 表示 log3 30 .
解:根据题意,得 b=log35,

∴log3 30

1 2

=

1

log3(3×10)=

1

(log33+log310)

2

2

= 1 [1+log3(2×5)] 2
= 1 (1+log32+log35) 2
= 1 (1+a+b). 2
能力提升 踮起脚,抓得住!

8.设 1 ? 1 =n,那么 n log 2 3 log5 3

()

A.(2,3) 答案:A

B.(1,2)

C.(-2,-1)

D.(-3,-2)

解析: 1 ? 1 =log32+log35=log310, log 2 3 log5 3

2=log39<log310<log327=3. ∴2<n<3.

9.若 lgx=a,lgy=b,则 lg
1
A. a-2b-2
2 1
C. a-2b-1
2

x -lg( y )2 的值为( ) 10 1
B. a-2b+2
2 1
D. a-2b+1
2

答案:B

解析:∵lg x -lg( y )2= 1 lgx-2lg y

10 2

10

= 1 lgx-2(lgy-lg10) 2

= 1 a-2(b-1) 2

= 1 a-2b+2. 2

10.已知

f(x)=

? ? ?

log 2 x, x2,

??? 2x? 3 ,

x ? (0,??), x ? (?1,0], x ?(0,??),则 f{f[f(-2-3)]}=______________. x ? (??,?1],

答案:-4

解析:∵-2- 3 <-1,

又∵x∈(-∞,-1)时,f(x)=- 2x? 3 ,

∴f(-2- 3 )=- 2?2? 3? 3 =- 1 . 4
∵- 1 ∈(-1,0], 4
而 x∈(-1,0)时,f(x)=x2,

∴f[f(-2- 3 )]=f(- 1 )=(- 1 )2= 1 >0. 4 4 16
而 x∈(0,+∞)时,f(x)=log2x,

∴f{f[f(-2-3)]}=f( 1 )=log2 1 =-4.

16

16

11.已知 a、b、c 为△ABC 的三边,且关于 x 的方程 x2-2x+lg(c2-b2)-2lga+1=0 有等根,判断△ABC

的形状.

解:由题意可得Δ =0, 即 4-4[lg(c2-b2)-2lga+1]=0. ∴2lga=lg(c2-b2),lga2=lg(c2-b2).

∴a2=c2-b2, 即 a2+b2=c2.

根据勾股定理可得△ABC 是直角三角形.

12.已知 am=2,an=3,求 a3m-2n 的值. 解:∵am=2,an=3,

∴loga2=m,loga3=n.

∴a = a ? a 3m-2n

3loga2

?2

log

3 a

log

8 a

?loga9

8
a = loga 9

?

8

.

9

13.计算:

(1)log2

7 48

1 +log212- 2

log242;

(2)lg2·lg50-lg5·lgg4.

解:(1)log2 7 +log212- 1 log242

48

2

=log2[ 7 ·12· 1 ]

43

7? 6

=log2( 1 ) 2

?1
=log2 2 2

=-

1

.

2

(2)lg2·lg50-lg5·lgg4=lg2(1+lg5)-lg5(1+lg2)-2lg2 =lg2-lg5-2lg2 =-(lg2+lg5)=-1. 拓展应用 跳一跳,够得着!

1
14. 定 义 运 算 法 则 如 下 :a*b= a 2

?1
?b 3

,a ?

1
b=lga2-lg b 2

,M=2

1

*

8

,N=

2 ? 1 ,则

4 125

25

M+N=_________________. 答案:5

解析:M= 9 *

8

=

(

9

)

1 2

?

(

8

?1
)3

?

3

?

5

=4.

4 125 4 125 2 2

N= 2 ? 1 25

=lg2-lg 1 =lg10=1. 5
∴M+N=5.

15.如果 log2[log3(log4x)]=log3[log4(log2y)]=log4[log2(log3z)]=0,则 x+y-z 等于( )

A.70

B.71

C.89

D.90

答案:B

解析:log2[log3(log4x)]=0 ?log3(log4x)=1 ?log4x=3.
∴x=43=64.

同理,y=16,z=9.

∴x+y-z=71.

16.已知二次函数 f(x)=(lga)x2+2x+4lga 的最大值为 3,求 a 的值.

解:原函数式可化成 f(x)=lga(x+ 1 )2- 1 +4lga. lg a lg a

由已知,f(x)有最大值 3,

∴lga<0,并且- 1 +4lga=3, lg a

整理得 4(lga)2-3lga-1=0,
解得 lga=1,lga=- 1 . 4
∵lga<0,故取 lga=- 1 . 4

?1
∴a=10 4

? 4 1000

.

10


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