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高中数学对数运算性质—换底公式课件人教版必修1_图文

? 练习: 1.已知lg2=a , lg3=b , 请用a ,b 表示 lg12 . 3-102)的结果( )。 2.计算lg ( 10 3 A. 1 B. 2 C. 90 D.2+lg9 1.解:lg12 =lg(4×3) =lg4+lg3 =2lg2+lg3 =2a +b ? 2.解: lg ( 103-102) = lg [102( 10-1)] = lg(102× 9) =lg102+lg9 =2+lg9 7 3.计算: (1)lg 14 ? 2 lg ? lg 7 ? lg 18 3 解法一: 解法二: 7 lg 14 ? 2 lg ? lg 7 ? lg 18 3 7 2 ? lg14 ? lg( ) ? lg 7 ? lg18 3 14? 7 ? lg 7 2 ( ) ?18 3 ? lg1 ? 0 7 lg 14 ? 2 lg ? lg 7 ? lg 18 3 7 ? lg(2 ? 7) ? 2 lg 3 2 ? lg 7 ? lg(2 ? 3 ) ? lg 2 ? lg 7 ? 2(lg 7 ? lg 3) ? lg 7 ? (lg 2 ? 2 lg 3) ?0 提高练习: ⑴ 若 lg x ? lg a ? 2lg b ? 3lg c, 则 x ? ______ c3 2 ab ⑵ ⑶ 1 log 6 12 ? log 6 2 2 _____________ 1 2 2 的值为______ log 2 8 ? 4 3 ? log 2 8 ? 4 3 ? (一)复习 积、商、幂的对数运算法则: 如果 a > 0,a ? 1,M > 0, N > 0 有: loga (MN) ? loga M ? loga N (1) M loga ? loga M ? loga N (2) N n loga M ? nlog (3) a M(n ? R) 二、新课: logm N 1.对数换底公式: loga N ? logm a ( a > 0 ,a ? 1 ,m > 0 ,m ? 1,N>0) 证明: 设 logaN=x ,则 ax= N,两边取以m为底的对数: 王新敞 奎屯 新疆 王新敞 奎屯 新疆 王新敞 奎屯 新疆 logm a ? logm N ? x logm a ? logm N x 从而得: x ? ∴ logm N logm a logm N loga N ? logm a 2.两个常用的推论: ① logablogba=1 , loga b ? logb c ? logc a ? 1 n n log b ? log a b ( a, b > 0且均不为1) a ② m m 王新敞 奎屯 新疆 王新敞 奎屯 新疆 王新敞 奎屯 新疆 王新敞 奎屯 新疆 证: lg b lg a loga b ? logb a ? ? ?1 lg a lg b loga m n lg b n lg b n n b ? ? ? loga b m m lg a m lg a 三、讲解范例: 例1 求log89.log2732的值. 分析:利用换底公式统一底数: 一般情况下,可换成常用对数,也可根据真、底数的特 征,换成其它合适的底数. 例2 已知 log2 3 = a, log3 7 = b,用 a, b 表示 log42 56 1 解:因为log23 = a,则 ? log 3 2 , 又∵ a log3 7 = b, ∴ log 42 log3 56 log3 7 ? 3 ? log3 2 ab ? 3 56 ? ? ? log3 42 log3 7 ? log3 2 ? 1 ab ? b ? 1 例3计算:① 5 1? log 0.2 3 ② log 4 3 ? log 9 2 ? log 1 4 32 2 5 解:①原式 = 5log0.2 3 ? log 1 ? 1 ? 15 5 5 3 3 1 1 5 1 5 3 log 3 ? log 2 ? log 2 ? ? ? ②原式 = 2 3 2 2 2 4 4 4 2 5 5 例3设 x, y, z ? (0,??) 且 3x=4y=6z 1? 求证 1 1 1 ? ? x 2y z ; 2? 比较 3x,4 y,6 z 的大小 例3设 x, y, z ? (0,??) 且3x=4y=6z 1? 求证 1 1 1 ? ? x 2y z ; 2? 比较 3x,4 y,6 z 的大小 x y z x, y, z ? (0,??) k ?1 3 ? 4 ? 6 ? k 证明1?:设 ∵ ∴ 取对数得: x? lg k lg 3 , y? lg k lg 4 z ? , lg k lg 6 1 1 lg 3 lg 4 2 lg 3 ? lg 4 2 lg 3 ? 2 lg 2 lg 6 1 ∴ x ? 2 y ? lg k ? 2 lg k ? 2 lg k ? 2 lg k ? lg k ? z 例3设 x, y, z ? (0,??) 1? 求证 1 1 1 ? ? x 2y z 且 3x ? 4 y ? 6 z ; 2? 比较 3x,4 y,6 z 的大小 证明1?:设3x ? 4 y ? 6 z ? k ∵x, y, z ? (0,??) ∴ 取对数得: x ? ∴ ? ? lg k lg 3 k ?1 ,y ? lg k lg 4 , z ? lg 6 lg k 1 1 lg 3 lg 4 2 lg 3 ? lg 4 2 lg 3 ? 2 lg 2 lg 6 1 ? ? ? ? ? ? ? x 2 y lg k 2 lg k 2 lg k 2 lg k lg k z lg k lg 3 4 81 ? 0 3x ? 4 y ? ( ? ) lg k ? lg 6

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