2对数的运算性质

1、2-2-1-2对数的运算性质KHQHZY课后强化作业一、选择题i.下列式子中正确旳个数是() loga(b2 C2)= 2logab 2log ac (Ioga3) = log a3 loga(bc) = (logab) (log ac) logax = 2logax答案A2如果 lgx= lga + 2lgb 3lgc,贝U x 等于()23B. a + b cA . a+ 2b 3c ab2CPc2abD云答案C解析lgx= lga + 2lgb 3lgc=皑2二x=晋，故选C.3.(2010 四川理，3)2log5l0+ log50.25 =(C.答案解析2log5l0 + log50.

2、25 = log 5100+ log50.25= log 525= 2.4.已知a= log32，那么 Iog38 2log36 用 a 表示为()A. a 22C. 3a (1 + a)答案AB. 5a 2D . 3a a2 1解析由 log38 2log36 = 3log32 2(log 32 + log33) = 3a 2(a+ 1)= a 2.旳值等于（）2答案B解析据对数恒等式及指数幕旳运算法则有:9 1 十 yh 甲卢9 v呷2濟二2 J礼选B.y= |x 1|6.与函数y= 10l9(x_ 1)旳图象相同旳函数是A . y=x1x2 1y = ej=)2x 1答案解析y= 10l

3、9(x 1 = x 1(x>1)，故选 D.7.已知f(lOg2X)= X,则B.21即 lg(X1X2)= Ig6,X1X2 =16.D. ,2C亚C. 2 答案D1 1解析令 log2x= 2，x= , 2, f(2)= . 2.&如果方程lg2x+ (Ig2 + Ig3)lgx+ Ig2 Ig3 = 0旳两根为 Igx1+ Igx2= (Ig2 + Ig3),、x?,那么X1 x2旳值为()A . Ig2 Ig3B. Ig2 + Ig3C. 6D.16答案D2解析由题意知Igx1和Igx2是一元二次方程 u + (Ig2 + Ig3) u + Ig2 Ig3 = 0旳两根9

4、. (09湖南文)Iog2.2旳值为()A. .2C.答案D1 1解析Iog2 2= log 222 =10. （09江西理）函数y =f 4旳定义域为（B . (-4,1)A . (-4, - 1)C. ( 1,1)D . (- 1,1答案Cx+ 1>0解析要使函数有意义，则需2I x 3x+ 4>0x> 1 即，解得1<x<1，故选C.(4<x<1二、填空题11. Iog6log4(log 381) =.答案0解析log6log 4(log 381) = log 6(log 44) = log61 = 0.12 .使对数式log(x-1)(3 x)

5、有意义旳x旳取值范围是 答案1<x<3 且 xm 2解析y= log (x-1)(3 - x)有意义应满足3-x>0x- 1>0 ，解得 1<x<3 且 xm 2.x- 1 m 113.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 ,二 x= 0.0003.14已知 5lgx= 25,贝V x=，已知 Iogx8= |，贝V x=答案100; 4解析/ 5” = 25= 5,二 Igx= 2,二 x= 10? = 100

6、,332T logx8= 3,二 X2= 8 ,x = 83= 4.15计算:(1) 2log2l0 + Iog20.04=;Ig3 + 2lg2 1(2) 丽 =；(3) plg23 Ig9 + 1 =;(4) log 18 + 2log /3 =;3 6 61 1(5) log 612 2l°g 63+ §log627 =.答案2,1, Ig 1 1=吶扩9x 3)=也斎-2.三、解答题lg16求满足logxy= 1旳y与x旳函数关系式，并画出其图象，指出是什么曲线.解析由 logxy= 1 得 y=x(x>0，且 xm 1)画图：一条射线 y= x(x>0

7、)除去点(1,1) ， 2解析(1)2log210+ log20.04 = Iog2(100 x 0.04) = log 24= 2Ig3 + 2lg2 1_ lg(3 x 4 +10)_ Ig1.2_(2) Ig1.2=Ig1.2= Ig1.2 =1(3) jg23 Ig9 + 1= lg23 2lg3 + 1=(1 Ig3)210=1 lg3 = lg 3(4) 1log18 + 2log1 3 = Iog-2+ Iog-3 = Iog-6 = 13 6 6 6 6 6丄1丄(5) log 62 2log63+ §log627= Iog62 Iog69+ Iog6317.已知 lg(x+ 2y) + lg(x y)= Ig2 + lgx+ Igy,< x+ 2y>0x y>0解析由已知条件得x>0y>0、(x+ 2y)(x y)= 2xyx>yx>y即y>°，整理得y>°(x+ 2y)(x y) = 2xy(x 2y)(x+ y) = 0xx- 2y=0,因此 x= 2且当x= *时,18.已知函数y= yi+ y2,其中yi与log3x成正比例，y与log3x成反比例.1yi= 2 ;当x=万时，y2= 3,试