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【课堂新坐标】高中数学苏教版必修1练习:3.2.1 第2课时 对数的运算性质

学业分层测评 (建议用时:45 分钟) 学业达标] 一、填空题 1.下列式子中成立的是(假定各式均有意义)________.(填序号) ①loga x· loga y=loga(x+y); ②(loga x)n=nloga x; ③ ④ loga x n =loga x; n loga x =loga x-loga y. loga y 【解析】 根据对数的运算性质知,③正确. 【答案】 ③ 1 1 2.设 7a=8b=k,且 + =1,则 k=________. a b 【解析】 ∵7a=k,∴a=log7k,∵8b=k,∴b=log8k. 1 1 ∴ + =logk7+logk8=logk56=1.∴k=56. a b 【答案】 56 16 3.已知 a2= (a>0),则 log2 a=________. 81 3 16 4 【解析】 由 a2= (a>0),得 a= , 81 9 2?2 4 所以 log2 =log2? 3 =2. ? 9 3 3 ? 【答案】 2 4.lg x1 与 lg x2 是方程(lg x)2+(lg 2+lg 3)lg x+lg 2· lg 3=0 的两根,则 x1x2=________. 【解析】 由题意, lg x1, lg x2 是关于 lg x 的一元二次方程(lg x)2+(lg 2+lg 3)lg x+lg 2· lg 3=0 的两个根,则 x1,x2 是关于 x 的方程的两个根,由根与系数的关系,得 lg x1+lg x2=- (lg 2+lg 3), 1 1 即 lg (x1x2)=lg ,∴x1x2= . 6 6 【答案】 1 6 x? 3 ? y? 3 5.若 lg x-lg y=a,则 lg ? ?2? -lg ?2? =________. x? 【解析】 lg x-lg y=lg ? ?y?=a, 3 3 x ?3 ?y?3=lg x -lg y =lg ?x?3=3lg ?x?=3a. lg ? - lg ?2? ?2? ?y? ?y? 8 8 【答案】 3a 6.若 lg 2=a,lg 3=b,则 log5 12 等于________. lg 12 lg 3+2lg 2 b+2a 【解析】 log5 12= = = . lg 5 1-lg 2 1-a 【答案】 b+2a 1-a 7.里氏震级 M 的计算公式为:M=lg A-lg A0,其中 A 是测震仪记录的地震曲线的最 大振幅,A0 是相应的标准地震的振幅.假设在一次地震中,测震仪记录的最大振幅是 1 000, 此时标准地震的振幅为 0.001,则此次地震的震级为________级;9 级地震的最大振幅是 5 级地震最大振幅的________倍. 【解析】 由 M=lg A-lg A0 知,M=lg 1 000-lg 0.001=6,所以此次地震的级数为 6 级.设 9 级地震的最大振幅为 A1,5 级地震的最大振幅为 A2,则 lg A1 =lg A1-lg A2=(lg A1 A2 A1 -lg A0)-(lg A2-lg A0)=9-5=4.所以 =104=10 000.所以 9 级地震的最大振幅是 5 级地震 A2 的最大振幅的 10 000 倍. 【答案】 6 10 000 1-x 8.(2016· 锦州高一检测)已知函数 f (x)=lg ,若 f (a)=b,则 f (-a)=________. 1+x 1-x 【解析】 因为 f (x)=lg , 1+x 1-a 所以 f (a)=lg =b, 1+a 1+a ?1-a?-1 所以 f (-a)=lg =lg? ? 1-a ?1+a? =-b. 【答案】 -b 二、解答题 9.计算: 7 (1)log5 35-2log5 +log5 7-log5 1.8; 3 lg (2) 27+lg 8-lg lg 1.2 1 000 ; (3)(lg 5)2+lg 2· lg 50. 9 【解】 (1)原式=log5(5× 7)-2(log5 7-log5 3)+log5 7-log5 =log5 5+log5 7-2log5 7 5 +2log5 3+log5 7-2log5 3+log5 5=2log5 5=2. 3 3 lg 3+3lg 2- 2 2 (2)原式= lg 3+2lg 2-1 = 3lg 3+6lg 2-3 3 = . 2 +2lg 2- (3)原式=(lg 5)2+lg 2· (lg 2+2lg 5) =(lg 5)2+2lg 5· lg 2+(lg 2)2 =(lg 5+lg 2)2=1. 10.(1)已知 10a=2,10b=3,求 1002a b; - 8 50 (2)设 a=lg 2,b=lg 7,用 a,b 表示 lg ,lg . 7 49 【解】 (1)∵10a=2, ∴lg 2=a, 又∵10b=3,∴lg 3=b. 8 (2)lg =lg 23-lg 7=3lg 2-lg 7=3a-b. 7 lg 50 =lg (2× 52)-lg (72)=lg 2+2lg 5-2lg 7 49 =lg 2+2(1-lg 2)-2lg 7 =2-a-2b. 能力提升] 1.化简: 【解析】 2 2 1 -4log2 3+4+log2 =________. 3 2 2 -4log2 3+4= - 2 3- 2 =2-log2 3. ∴原式=2-log2 3+log2 3 1=2-2log2 3. 【答案】 2-2log2 3 1 2.设 a 表示 的小数部分,则 log2a(2a+1)的值是________. 3- 5 【解析】 3+ 5 1 = , 4 3- 5 3+ 5 5-1 可得 a= -1= . 4 4 【答案】 -1 3.若 a,b 是方程 2(lg x)2-lg x4+1=0 的两个实根,则 lg(ab)· (logab+logba)的值为 ______

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