1、对数与对数函数测试题一、 选择题:1已知35= A,且= 2,则A的值是( )(A)15 (B) (C) (D)2252已知a0,且10= lg(10x)lg,则x的值是( )(A)1 (B)0 (C)1 (D)23若x,x是方程lgx (lg3lg2)lg3lg2 = 0的两根,则xx的值是( )(A)lg3lg2 (B)lg6 (C)6 (D)4若log(a1)log2a0,那么a的取值范围是( )(A)(0,1) (B)(0,) (C)(,1) (D)(1,)5 已知x =,则x的值属于区间( )(A)(2,1) (B)(1,2) (C)(3,2) (D)(2,3) 6已知lga,lgb
2、是方程2x4x1 = 0的两个根,则(lg)的值是( )(A)4 (B)3 (C)2 (D)17设a,b,cR,且3= 4= 6,则( )(A)= (B)= (C)= (D)=8已知函数y = log(ax2x1)的值域为R,则实数a的取值范围是( )(A)0a1 (B)0a1 (C)a1 (D)a19已知lg20.3010,且a = 285的位数是M,则M为( )(A)20 (B)19 (C)21 (D)2210若log log( logx) = 0,则x为( )(A) (B) (C) (D) 11若0a1,函数y = log1()在定义域上是( )(A)增函数且y0 (B)增函数且y0 (
3、C)减函数且y0 (D)减函数且y012已知不等式log(1)0的解集是(,2),则a的取值范围是( )(A)0a (B)a1 (C)0a1 (D)a1二、 填空题13若lg2 = a,lg3 = b,则lg=_14已知a = log0.8,b = log0.9,c = 1.1,则a,b,c的大小关系是_15log(32) = _16设函数= 2(x0)的反函数为y =,则函数y =的定义域为_三、 解答题17已知lgx = a,lgy = b,lgz = c,且有abc = 0,求xyx的值18要使方程xpxq = 0的两根a、b满足lg(ab) = lgalgb,试确定p和q应满足的关系1
4、9设a,b为正数,且a2ab9b= 0,求lg(aab6b)lg(a4ab15b)的值20已知log log( logx) = log log( logy) = log log( logz) = 0,试比较x、y、z的大小21已知a1,= log(aa) 求的定义域、值域;判断函数的单调性 ,并证明;解不等式:22已知= loga2(ab)b1,其中a0,b0,求使0的x的取值范围参考答案:一、选择题:1(B)2(B) 3(D)4(C)5(D)6(C)7(B)8(A) 9(A)10(D)11(C)12(D)提示:135= A,a = logA,b = logA,= log3log5 = log
5、15 = 2,A =,故选(B) 210= lg(10x)lg= lg(10x) = lg10 = 1,所以 x = 0,故选(B) 3由lg xlg x=(lg3lg2),即lg xx= lg,所以xx=,故选(D)4当a1时,a12a,所以0a1,又log2a0,2a1,即a,综合得a1,所以选(C)5x = loglog= log() = log= log10,91027, 2log103,故选(D)6由已知lgalgb = 2,lgalgb =,又(lg)= (lgalgb)= (lgalgb)4lgalgb = 2,故选(C)7设3= 4= 6= k,则a = logk,b= log
6、k,c = logk,从而= log6 = log3log4 =,故=,所以选(B)8由函数y = log(ax2x1)的值域为R,则函数u(x) = ax2x1应取遍所有正实数,当a = 0时,u(x) = 2x1在x时能取遍所有正实数;当a0时,必有0a1所以0a1,故选(A) 9lga = lg(285) = 7lg211lg810lg5 = 7 lg2113lg210(lg10lg2) = 30lg21019.03,a = 10,即a有20位,也就是M = 20,故选(A)10由于log( logx) = 1,则logx = 3,所以x = 8,因此 x= 8=,故选(D)11根据u(
7、x) = ()为减函数,而()0,即1()1,所以y = log1()在定义域上是减函数且y0,故选(C)12由x2知,11,所以a1,故选(D)二、填空题13ab 14bac 152 16x1提示:13lg=lg(23) =( lg23lg3) =ab140a = log0.8log0.7 = 1,b = log0.90,c = 1.11.1= 1,故bac1532= (1),而(1)(1) = 1,即1= (1),log(32) =log(1)=216= logx (0x1,y =的定义域为02x11,即x1为所求函数的定义域二、 解答题17由lgx = a,lgy = b,lgz = c
8、,得x = 10,y = 10,z = 10,所以xyx=10=10= 10=18由已知得, 又lg(ab) = lgalgb,即ab = ab,再注意到a0,b0,可得p = q0,所以p和q满足的关系式为pq = 0且q019由a2ab9b= 0,得()2()9 = 0,令= x0,x2x9 = 0,解得x =1,(舍去负根),且x= 2x9,lg(aab6b)lg(a4ab15b) = lg= lg= lg= lg= lg= lg= lg=20由log log( logx) = 0得,log( logx)= 1,logx =,即x = 2;由log log( logy) = 0得,log
9、( logy) = 1,logy =,即y =3;由log log( logz) = 0得,log( logz) = 1,logz =,即z = 5y =3= 3= 9,x = 2= 2= 8,yx,又x = 2= 2= 32,z = 5= 5= 25,xz故yxz21为使函数有意义,需满足aa0,即aa,当注意到a1时,所求函数的定义域为(,1),又log(aa)loga = 1,故所求函数的值域为(,1)设xx1,则aaaa,所以= log(aa)log(aa)0,即所以函数为减函数 易求得的反函数为= log(aa) (x1),由,得log(aa)log(aa),aa,即x2x,解此不等式,得1x2,再注意到函数的定义域时,故原不等式的解为1x122要使0,因为对数函数y = logx是减函数,须使a2(ab)b11,即a2(ab)b0,即a2(ab)b2b,(ab)2b,又a0,b0,abb,即a(1)b,所以()1当ab0时,xlog(1);当a = b0时,xR;当ba0时,xlog(1)综上所述,使0的x的取值范围是:当ab0时,xlog(1);当a = b0时,xR;当ba0时,xlog(1)