高一对数函数运算法则的证明

2个回答

  • 高一对数函数运算法则

    1、a^(log(a)(b))=b (对数恒等式)

    2、log(a)(a^b)=b

    3、log(a)(MN)=log(a)(M)+log(a)(N);

    4、log(a)(M÷N)=log(a)(M)-log(a)(N);

    5、log(a)(M^n)=nlog(a)(M)

    6、log(a^n)M=1/nlog(a)(M)

    证明:

    1、因为n=log(a)(b),代入则a^n=b,即a^(log(a)(b))=b.

    2、因为a^b=a^b

    令t=a^b

    所以a^b=t,b=log(a)(t)=log(a)(a^b)

    3、MN=M×N

    由基本性质1(换掉M和N)

    a^[log(a)(MN)] = a^[log(a)(M)]×a^[log(a)(N)] =(M)*(N)

    由指数的性质

    a^[log(a)(MN)] = a^{[log(a)(M)] + [log(a)(N)]}

    两种方法只是性质不同,采用方法依实际情况而定

    又因为指数函数是单调函数,所以

    log(a)(MN) = log(a)(M) + log(a)(N)

    4、与(3)类似处理

    MN=M÷N

    由基本性质1(换掉M和N)

    a^[log(a)(M÷N)] = a^[log(a)(M)]÷a^[log(a)(N)]

    由指数的性质

    a^[log(a)(M÷N)] = a^{[log(a)(M)] - [log(a)(N)]}

    又因为指数函数是单调函数,所以

    log(a)(M÷N) = log(a)(M) - log(a)(N)

    5、与(3)类似处理

    M^n=M^n

    由基本性质1(换掉M)

    a^[log(a)(M^n)] = {a^[log(a)(M)]}^n

    由指数的性质

    a^[log(a)(M^n)] = a^{[log(a)(M)]*n}

    又因为指数函数是单调函数,所以

    log(a)(M^n)=nlog(a)(M)

    基本性质4推广

    log(a^n)(b^m)=m/n*[log(a)(b)]

    推导如下:

    由换底公式(换底公式见下面)[lnx是log(e)(x),e称作自然对数的底]

    log(a^n)(b^m)=ln(b^m)÷ln(a^n)

    换底公式的推导:

    设e^x=b^m,e^y=a^n

    则log(a^n)(b^m)=log(e^y)(e^x)=x/y

    x=ln(b^m),y=ln(a^n)

    得:log(a^n)(b^m)=ln(b^m)÷ln(a^n)

    由基本性质4可得

    log(a^n)(b^m) = [m×ln(b)]÷[n×ln(a)] = (m÷n)×{[ln(b)]÷[ln(a)]}

    再由换底公式

    log(a^n)(b^m)=m÷n×[log(a)(b)]

    例如:log(8)27=log(2³)3³=log(2)3

    再如:log(√2)√5=log(2)5.