第1个回答 2011-10-14
2.2.1.2
一、选择题
1.下列式子中正确的个数是( )
①loga(b2-c2)=2logab-2logac
②(loga3)2=loga32
③loga(bc)=(logab)•(logac)
④logax2=2logax
A.0 B.1 C.2 D.3
[答案] A
2.如果lgx=lga+2lgb-3lgc,则x等于( )
A.a+2b-3c B.a+b2-c3
C.ab2c3 D.2ab3c
[答案] C
[解析] lgx=lga+2lgb-3lgc=lgab2c3,
∴x=ab2c3,故选C.
3.(2010•四川理,3)2log510+log50.25=( )
A.0 B.1
C.2 D.4
[答案] C
[解析] 2log510+log50.25=log5100+log50.25=log525=2.
4.已知a=log32,那么log38-2log36用a表示为( )
A.a-2 B.5a-2
C.3a-(1+a)2 D.3a-a2-1
[答案] A
[解析] 由log38-2log36=3log32-2(log32+log33)=3a-2(a+1)=a-2.
5. 的值等于( )
A.2+5 B.25
C.2+52 D.1+52
[答案] B
[解析] 据对数恒等式及指数幂的运算法则有:
6.与函数y=10lg(x-1)的图象相同的函数是( )
A.y=x-1 B.y=|x-1|
C.y=x2-1x+1 D.y=(x-1x-1)2
[答案] D
[解析] y=10lg(x-1)=x-1(x>1),故选D.
7.已知f(log2x)=x,则f(12)=( )
A.14 B.12
C.22 D.2
[答案] D
[解析] 令log2x=12,∴x=2,∴f(12)=2.
8.如果方程lg2x+(lg2+lg3)lgx+lg2•lg3=0的两根为x1、x2,那么x1•x2的值为( )
A.lg2•lg3 B.lg2+lg3
C.-6 D.16
[答案] D
[解析] 由题意知lgx1和lgx2是一元二次方程u2+(lg2+lg3)u+lg2•lg3=0的两根
∴lgx1+lgx2=-(lg2+lg3),
即lg(x1x2)=lg16,∴x1x2=16.
9.(09•湖南文)log22的值为( )
A.-2 B.2
C.-12 D.12
[答案] D
[解析] log22=log2212=12.
10.(09•江西理)函数y=ln(x+1)-x2-3x+4的定义域为( )
A.(-4,-1) B.(-4,1)
C.(-1,1) D.(-1,1]
[答案] C
[解析] 要使函数有意义,则需x+1>0-x2-3x+4>0,
即x>-1-4<x<1,解得-1<x<1,故选C.
二、填空题
11.log6[log4(log381)]=________.
[答案] 0
[解析] log6[log4(log381)]=log6(log44)=log61=0.
12.使对数式log(x-1)(3-x)有意义的x的取值范围是________.
[答案] 1<x<3且x≠2
[解析] y=log(x-1)(3-x)有意义应满足
3-x>0x-1>0x-1≠1,解得1<x<3且x≠2.
13.已知lg3=0.4771,lgx=-3.5229,则x=________.
[答案] 0.0003
[解析] ∵lgx=-3.5229=-4+0.4771
=-4+lg3=lg0.0003,∴x=0.0003.
14.已知5lgx=25,则x=________,已知logx8=32,则x=________.
[答案] 100;4
[解析] ∵5lgx=25=52,∴lgx=2,∴x=102=100,
∵logx8=32,∴x32=8,∴x=823=4.
15.计算:
(1)2log210+log20.04=________;
(2)lg3+2lg2-1lg1.2=________;
(3)lg23-lg9+1=________;
(4)13log168+2log163=________;
(5)log6112-2log63+13log627=________.
[答案] 2,1,lg103,-1,-2
[解析] (1)2log210+log20.04=log2(100×0.04)=log24=2
(2)lg3+2lg2-1lg1.2=lg(3×4÷10)lg1.2=lg1.2lg1.2=1
(3)lg23-lg9+1=lg23-2lg3+1=(1-lg3)2
=1-lg3=lg103
(4)13log168+2log163=log162+log163=log166=-1
(5)log6112-2log63+13log627=log6112-log69+log63
=log6(112×19×3)=log6136=-2.
三、解答题lg
16.求满足logxy=1的y与x的函数关系式,并画出其图象,指出是什么曲线.
[解析] 由logxy=1得y=x(x>0,且x≠1)
画图:一条射线y=x(x>0)除去点(1,1).
17.已知lg(x+2y)+lg(x-y)=lg2+lgx+lgy,求xy的值.
[解析] 由已知条件得x+2y>0x-y>0x>0y>0(x+2y)(x-y)=2xy
即x>yy>0(x+2y)(x-y)=2xy,整理得x>yy>0(x-2y)(x+y)=0
∴x-2y=0,因此xy=2.
18.已知函数y=y1+y2,其中y1与log3x成正比例,y2与log3x成反比例.且当x=19时,y1=2;当x=127时,y2=-3,试确定函数y的具体表达式.
[解析] 设y1=klog3x,y2=mlog3x,
∴当x=19时,klog319=2,∴k=-1
当x=127时,mlog3127=-3,∴m=9
∴y=y1+y2=-log3x+9log3x.