高一数学必修1 对数的运算性质 ppt.ppt

对数的运算性质奎屯王新敞新疆奎屯王新敞新疆奎屯王新敞新疆奎屯王新敞新疆奎屯王新敞新疆1）对数的定义：，请同学们回顾一下上节课我们所学习的内容。一般地，如果的b次幂等于N,就是1,0aaaNabbNalog，那么数b叫做以a为底N的对数，记作.2)指数式与对数式的互化bNNaablog3）重要结论用公式①负数与零没有对数；②③对数恒等式1log,01logaaa.logNaNa奎屯王新敞新疆奎屯王新敞新疆奎屯王新敞新疆奎屯王新敞新疆奎屯王新敞新疆，指数的运算性质：mnmnmnnmnmnmnmnmaaRnmaaRnmaaaaaa)4();,())(3();,()2(;)1(M（任意给出）603.14162023150.3567N（任意给出）322.71830.785656.89lgMlgNlgM+lgNlgM-lgNlg（MN）lglgMlgNlg（M+N）lg（M-N）NMlglgnmlgmnlgNMM（任意给出）603.14162023150.3567N（任意给出）322.71830.785656.89lgM1.77820.49725.3060-0.4477lgN1.50510.4343-0.10481.7550lgM+lgN3.28330.93155.20121.3073lgM-lgN0.27310.06295.4108-2.2027lg（MN）3.28330.93155.20121.3073lg0.27310.06295.4108-2.2027lgMlgN2.67630.2159-0.5561-0.78561.00441.1448-50.6308-0.2551lg（M+N）1.96380.76795.30601.7578lg（M-N）1.4471-0.37345.3060出错信息56.90081.35154.1684-25.469756.90081.35154.1684-25.4697NMlglgnmlgmnlgNMM（任意给出）603.14162023150.3567N（任意给出）322.71830.785656.89lgM1.77820.49725.3060-0.4477lgN1.50510.4343-0.10481.7550lgM+lgN3.28330.93155.20121.3073lgM-lgN0.27310.06295.4108-2.2027lg（MN）3.28330.93155.20121.3073lg0.27310.06295.4108-2.2027lgMlgN2.67630.2159-0.5561-0.78561.00441.448-50.6308-0.2551lg（M+N）1.96380.76795.30601.7578lg（M-N）1.4471-0.37345.3060出错信息56.90081.35154.1684-25.469756.90081.35154.1684-25.4697NMlglgnmlgmnlgNMM（任意给出）603.14162023150.3567N（任意给出）322.71830.785656.89lgM1.77820.49725.3060-0.4477lgN1.50510.4343-0.10481.7550lgM+lgN3.28330.93155.20121.3073lgM-lgN0.27310.06295.4108-2.2027lg（MN）3.28330.93155.20121.3073lg0.27310.06295.4108-2.2027lgMlgN2.67630.2159-0.5561-0.78561.00441.448-50.6308-0.2551lg（M+N）1.96380.76795.30601.7578lg（M-N）1.4471-0.37345.3060出错信息56.90081.35154.1684-25.469756.90081.35154.1684-25.4697NMlglgnmlgmnlgNMM（任意给出）603.14162023150.3567N（任意给出）322.71830.785656.89lgM1.77820.49725.3060-0.4477lgN1.50510.4343-0.10481.7550lgM+lgN3.28330.93155.20121.3073lgM-lgN0.27310.06295.4108-2.2027lg（MN）3.28330.93155.20121.3073lg0.27310.06295.4108-2.2027lgMlgN2.67630.2159-0.5561-0.78561.00441.448-50.6308-0.2551lg（M+N）1.96380.76795.30601.7578lg（M-N）1.4471-0.37345.3060出错信息56.90081.35154.1684-25.469756.90081.35154.1684-25.4697NMlglgnmlgmnlgNMM（任意给出）603.14162023150.3567N（任意给出）322.71830.785656.89lgM1.77820.49725.3060-0.4477lgN1.50510.4343-0.10481.7550lgM+lgN3.28330.93155.20121.3073lgM-lgN0.27310.06295.4108-2.2027lg（MN）3.28330.93155.20121.3073lg0.27310.06295.4108-2.2027lgMlgN2.67630.2159-0.5561-0.78561.00441.448-50.6308-0.2551lg（M+N）1.96380.76795.30601.7578lg（M-N）1.4471-0.37345.3060出错信息56.90081.35154.1684-25.469756.90081.35154.1684-25.4697NMlglgnmlgmnlgNM对数的运算性质两个正数的积的对数等于这两个正数的对数和两个正数的商的对数等于这两个正数的对数差logloglogaaaMNMN⑴logloglogaaaMMNN(2)loglog()naaMnMnR(3)语言表达:一个正数的n次方的对数等于这个正数的对数n倍如果a0，a1，M0，N0有：证明：②设,logpMa,logqNa由对数的定义可以得：,paMqaN∴qpaaqpaqpNMalog即证得NMlogloglogaaaMMNN证明:aaaMloglogMlogNN证明：设,logpMa由对数的定义可以得：,paM∴npnaMnpMnalog即证得naalogMnlogM(nR)loglognaaMnM证明:对数运算公式几个注意点：1)简易语言表达:“积的对数=对数的和”……2)真数的取值必须是(0,＋∞)3）有时公式可以可逆4）5）log()aMN≠loglogaaMNlog()aMN≠loglogaaMNnma)(lognamlog≠例1计算（1）（2）)42(log7525lg100讲解范例解:)42(log752522log724log522log1422log=5+14=19解:21lg1052lg105255lg100例2计算（1）（2）50lg)2(lg)5(lg2解(1)(2)原式=18lg7lg37lg214lg)15)(lg2(lg)5(lg50lg)2(lg)5(lg222lg5lg)2(lg)5(lg22lg)2lg5)(lg5(lg12lg5lg)23lg(7lg)3lg7(lg2)72lg(22lg3lg27lg3lg27lg27lg2lg0法二:原式=18lg7lg)37lg(14lg201lg18)37(714lg2练习（1）（4）（3）（2）1.求下列各式的值：15log5log332lg5lg31log3log553log6log2236log2)25lg()313(log5155log32log2110lg11log50133log12.用lgｘ，lgｙ，lgｚ表示下列各式：练习（1）（4）（3）（2）)lg(xyzzxy2lgzxy3lg＝lgｘ＋２lgｙ－lgｚ；zyx2lg＝lgｘ＋lgｙ＋lgｚ；＝lgｘ＋３lgｙ－21lgｚ；zyxlglg2lg21