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高中数学第二章指数函数、对数函数和幂函数2.2.2换底公式练习湘教版必修1

2.2.2 换底公式 [学习目标] 1.能记住换底公式, 并会证明换底公式.2.会利用换底公式解决一些对数式的化 简、求值、证明问题.3.能综合利用对数的相关知识解决问题. [预习导引] 1.对数的换底公式 logcN 换底公式:logaN= (a>0,a≠1,c>0,c≠1,N>0). logca 最常用的换底公式是 logaN= 2.换底公式的两个重要推论 (1)loga b = logab. (2)logab= 1 . logba m n lg N ln N 和 logaN= . lg a ln a n m 解决学生疑难点 ___________________________________________ __________________________________________________________ __________________________________________________________ __________________________________________________________ 要点一 利用换底公式求值或化简 例 1 求解下列各题: lg 2 (1)化简(log43+log83) ; lg 3 (2)已知 log1227=a,求 log616 的值. 解 (1)方法一 原式=? =? = ?lg 3+lg ?lg 4 lg 3?lg 2 ? 8?lg 3 ? lg 3 + lg 3 ?·lg 2 ? ?2lg 2 3lg 2? lg 3 lg 3 lg 2 lg 3 lg 2 1 1 5 · + · = + = . 2lg 2 lg 3 3lg 2 lg 3 2 3 6 2 3 方法二 原式=(log2 3+log2 3)·log32 1 1 5 5 ?1 ? =? log23+ log23?·log32= log23·log32= . 3 6 6 ?2 ? 3lg 3 (2)方法一 由 log1227=a,得 =a, 2lg 2+lg 3 3-a ∴lg 2= lg 3. 2a lg 16 4lg 2 ∴log616= = = lg 6 lg 2+lg 3 3 3-a 4× 2a = 3-a 1+ 2a -a . 3+a 方法二 由于 log1227=log123 =3log123=a, ∴log123= . 3 3 3 于是 log312= ,即 1+2log32= . a a a 3-a 因此 log32= . 2a 而 log616=4log62= 4 4 = = log26 1+log23 4 = = 1 2a 1+ 1+ log32 3-a 4 -a . 3+a 故 log616= -a . 3+a 规律方法 1.利用对数的换底公式计算化简时,通常有以下几种思路: 一是先依照运算性质:利用对数的运算法则及性质进行部分运算,最后再换成同一底. 二是一次性地统一换为常用对数,再化简、通分、求值. 三是将式子中的对数的底数及真数改写为幂的形式,然后利用变形 loga b = logab. 对出现的对数进行化简,当底数和真数都较小时,容易发现它们之间的关系,然后再借助对 数的运算法则求值. 2.对于换底公式,除了正用以外,也要注意其逆用以及变形应用. 跟踪演练 1 (1)求值:log89·log2732. (2)已知 log23=a,log37=b,试用 a,b 表示 log1456. lg 9 lg 32 2lg 3 5lg 2 10 解 (1)方法一 log89·log2732= · = · = . lg 8 lg 27 3lg 2 3lg 3 9 方法二 log89·log2732=log2 3 ·log3 2 2 5 10 = log23· log32= . 3 3 9 log27 log27 (2)∵log23=a,∴log37= = =b. log23 a 2 3 2 3 5 m n n m ∴log27=ab. log256 log28+log27 3+log27 3+ab ∴log1456= = = = . log214 log22+log27 1+log27 1+ab 要点二 利用对数的换底公式证明等式 2 1 2 a b c 例 2 已知 a,b,c 均为正数,3 =4 =6 ,求证: + = . a b c 证明 不妨设 3 =4 =6 =m,则 m>0 且 m≠1, 于是 a=log3m,b=log4m,c=log6m. 1 1 1 则由换底公式可得 =logm3, =logm4, =logm6, a b c a b c 2 1 2 于是 + =2logm3+logm4=logm(3 ×4) a b 2 =logm36=2logm6= . c 因此等式成立. 规律方法 1.在已知条件中出现幂值相等的形式时,通常可以设出幂值的结果,然后将指数 式转化为对数式,然后结合对数的换底公式、运算法则等进行化简和变形. 2. 由于对数的运算法则都是针对同底数的对数才能成立的, 因此变换底数是解决对数式证明 1 问题的重要环节,当出现的对数的底数不同,但真数相同时,可利用性质 logab= 进行 logba 变换. 跟踪演练 2 已知 2 =5 =10,求证:m+n=mn. 证明 由已知可得 m=log210,n=log510, 1 1 因此 =lg 2, =lg 5, m n m n 1 1 于是 + =lg 2+lg 5=lg 10=1, m n 即 n+m =1,故 m+n=mn. mn 要点三 对数换底公式的综合应用 1 1 a b 例 3 (1)已知 11.2 =1 000,0.011 2 =1 000,求 - 的值; a b (2)设 logac,logbc 是方程 x -3x+1=0 的两根,求

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