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高中数学 2-2-1-2对数的运算性质课件 新人教A版必修1_图文

第二章 基本初等函数(Ⅰ) 第二章 2.2 对 数 函 数 第二章 2.2.1 对数与对数运算 第二章 第 2 课时 对数的运算性质 课前自主预习 名师辩误做答 方法警示探究 思路方法技巧 课堂基础巩固 探索延拓创新 课后强化作业 课前自主预习 温故知新 1.对数 logaN(a>0,且 a≠1)具有下列简单性质: (1) 零和负数 没有对数,即 N > 0; (2)1 的对数为 0 ,即 loga1= 0 ; (3)底的对数等于 1 ;即 logaa= 1 ; (4)logaab= b . 2.对数与指数间的关系:当 a>0,a≠1 时,ax=N?x = logaN . 3.指数的运算法则:a>0,b>0,r,s∈R, ar· as= ar+s , r-s a a÷ a= , r s rs (ar)s= a , (ab)r= arbr . 新课引入 一提到有理数、实数,总能让我们联想到它们的四则运 算,这样的运算在“数”的内容中是举足轻重的.从我们对 数的认识来看,每引入一种新的数,总要研究其内在的运算 性质和规律.对数也是数,并且是与指数幂有着密切关系的 数.我们认识了对数的概念和对数本身固有的性质,那么, 对数是如何引到运算中的呢?其具体的运算是怎样的呢? 自主预习 问题:回顾指数幂的运算性质,am· an=am+n,(ap)m=apm, am÷ an=am-n,我们已经知道指数式与对数式可以互化,那么, 它们在运算上有没有必然的联系呢? 探究:a>0 且 a≠1 时, (1)设 M=am,N=an,显然 M>0,N>0, 则 M· N=am· an=am+n. 根据对数的定义有 logaM=m,logaN=n, loga(MN)=m+n. 于是有 logaM+logaN=m+n=loga(MN). m M a (2)设 M=am,N=an,则 = n =am-n. N a 根据对数的定义有 logaM=m,logaN=n, M loga =m-n. N M 于是有 logaM-logaN=m-n=loga N . 总结:对数的加、减法有: loga(MN)=logaM+logaN , M loga = logaM-logaN . N 【归纳提升】 1.我们来研究 loga(MN)=logaM+logaN 中 M=N 时的情况,有 logaM2=2logaM. 类似地有 logaM3=logaM+logaM+logaM=3logaM. 那么 logaMn=logaM+logaM+…+loga=nlogaM. n个 对于运算性质(1)、(2)可以理解成真数的乘除表现为对数 的加减,真数的乘方表现为对数与真数指数的积,可以形象 地认为对数运算是一种“降级”运算. 2.对对数运算性质的理解与运用时需注意的两个问题: (1)对于上面的每一条运算性质,都要注意只有当式子中 所有的对数记号都有意义时, 等式才成立. 如 log2[(-3)· (-5)] 是存在的,但 log2( - 3) 与 log2( - 5) 均不存在,故不能写成 log2[(-3(· (-5)]=log2(-3)+log2(-5). (2)利用对数的运算性质可以把乘、除、乘方的运算转化 为对数的加、减、乘运算,反之亦然,这种运算的互化可简 化计算,加快计算速度. 通过以上所学,完成下列练习. (1)log62+log63=________. (2)log312-log34=________. (3)2log93=________. (4)log3(95×34)=________. 7 (5)lg 1 000=________. [答案] 1 1 1 3 14 7 思路方法技巧 命题方向 1 对数的运算性质 [例 1] 用 logax,logay,logaz 表示: 2 (1)loga(xy );(2)loga(x y);(3)loga 3 x yz2. [解析] (1)loga(xy2)=logax+logay2=logax+2logay; 1 y=logax+2logay; (2)loga(x y)=logax+loga 3 (3)loga x 1 x 1 1 2 yz2 = 3 loga yz2 = 3 (logax - loga(yz )) = 3 (logax - logay-2logaz). 用 logax、logay、logaz 表示下列各式: x (1)loga(x y ); (2)loga . yz 3 5 [解析] (1)loga(x3y5)=logax3+logay5 =3logax+5logay; x (2)loga =loga x-loga(yz) yz =logax -(logay+logaz) 1 =2logax-logay-logaz. 1 2 命题方向 2 运用对数的运算性质解题 [例 2] 7 计算:(1)lg14-2lg +lg7-lg18; 3 2lg2+lg3 (2) ; 2+lg0.36+2lg2 (3)lg25+lg2· lg50. [分析] 当对数的底数相同时,利用对数运算的性质, 将式子转化为只含一种或少数几种真数的形式再进行计算. [解析] lg(32×2) (1)方法一:原式=lg(2×7)-2(lg7-lg3)+lg7- =lg2+lg7-2lg7+2lg3+lg7-2lg3-lg2 =0. 72 方法二:原式=lg14-lg( ) +lg7-lg18 3 14×7 =lg 7 ?3?2×18 =lg1 =0. 2lg2+lg3 2lg2+lg3 1 (2)原式= = = . 2+lg36-2+2lg2 4lg2+2lg3 2 (3)原式=lg25+(1-lg5)(1+lg5) =lg25+1-lg25 =1. 规律总结:灵活动

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