课件编号13701528

4.2.1对数的运算性质- 课件(共30张PPT)

日期:2024-06-01 科目:数学 类型:高中课件 查看:39次 大小:931512Byte 来源:二一课件通
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4.2.1,对数,运算,性质,课件,30张
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4.2.1对的运算性质 新课导入 10000=104 ? ????????104=4 1000=103 ? lg103=3 10000×1000=107 ? ????????107=7 ? 104×103=107?lg107=7=4+3=????????104+????????103 ? 观察下面的运算,能发现什么规律? 新知探究 104×103=107?lg107=7=4+3=????????104+????????103 ? 能否从指数运算????????·????????得到相应的对数运算? ? 新知探究 设M=???????? ; N=???????? 则 m=????????????????????; ????=???????????????????? 又????????·????????=????????+???? =????????? 所以 m+n=????????????????(????????) 即????????????????(????????)=m+n=????????????????????+???????????????????? ? 新知探究 思考:类似地能否从指数运算????????÷????????得到相应的对数运算? ? 新知探究 思考:类似地能否从指数运算????????÷????????得到相应的对数运算? ? ????????????????????????=?????????????????????????????????????????. ? 新知探究 再思考:由(????????)????=????????????能得到相应的对数运算吗? ? 新知探究 再思考:由(????????)????=????????????能得到相应的对数运算吗? ? ????????????????????????=???????????????????????? ? 对数的运算性质 设????>????,且????≠????,????>????,????>????,则 ? ?????????????????????????=????????????????????+???????????????????? ????????????????????????=????????????????????????????????????????? ????????????????????????=???????????????????????? ? 学以致用 例1 计算: (1)log264×512; (2)lg0.0001; (3)log3581. ? 学以致用 例1 计算: (1)log264×512; (2)lg0.0001; (3)log3581. ? 解:(1)?log264×512=log264+log2512=6+9=15; (2)?lg0.0001=lg10?4=?4lg10=?4; (3)?log3581=log3345=45log33=45. ? 针对练习 练习1:计算 (1)log264×16; (2)log39×27; (3)log1215122; (4)log336?log312; (5)log759+log7935; (6)lg20+lg5. ? 针对练习 练习1:计算 (1)log264×16; (2)log39×27; (3)log1215122; (4)log336?log312; (5)log759+log7935; (6)lg20+lg5. ? 解:(1)log264×16=log264+log216=log226+log224 =6log22+4log22=6+4=10; (2)log3(9×27)=log39+log327=log332+log333 =2log33+3log33=2+3=5; (3)log12(1512)2=log12(129)2=log12(12)18=18log1212=18; ? 针对练习 练习1:计算 (1)log264×16; (2)log39×27; (3)log1215122; (4)log336?log312; (5)log759+log7935; (6)lg20+lg5. ? 解:(4)log336?log312=log34×9?log33×4 =log34+log39?log33+log34=log332?log33=1; (5)log759+log7935=(log75?log79)+(log79?log735) =log75?log79+log79?log77?log75 =?log77=?1; (6)lg20+lg5=lg(20×5)=lg102=2. ? 15 方法小结 16 学以致用 17 18 19 20 21 学以致用 例3 已知log23=????,log25=????,用????,????表示下列各数的值: (1)log230;(2)log259;(3)log231520. ? 学以致用 例3 已知log23=????,log25=????,用????,????表示下列各数的值: (1)log230;(2)log259;(3)log231520. ? 解:(1)log230=log22×3×5=log22+log23+log25=1+????+????; (2)log259=log25?log29=log25?log232 =log25?2log23=?????2????; (3)log231520=log21513?log22012=13log215?12log220 =13log23+log25?12log24+log25=????3?????6?1. ? 针对练习 练习3:用lg ????,lg ????,lg ????表示下列各式: (1)lg????????????; (2)lg????2????????3; (3)lg????2????????3; (4)lg?????12????????23. ? 针对练习 练习3:用lg ????,lg ????,lg ????表示下列各式: (1)lg????????????; (2)lg????2????????3; (3)lg????2????????3; (4)lg?????12 ... ...

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