求对数3个运算性质 的推导啊

1个回答

  • 对数的性质及推导

    用^表示乘方,用log(a)(b)表示以a为底,b的对数

    *表示乘号,/表示除号

    定义式:

    若a^n=b(a>0且a≠1)

    则n=log(a)(b)

    基本性质:

    1.a^(log(a)(b))=b

    2.log(a)(MN)=log(a)(M)+log(a)(N);

    3.log(a)(M/N)=log(a)(M)-log(a)(N);

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

    推导

    1.这个就不用推了吧,直接由定义式可得(把定义式中的[n=log(a)(b)]带入a^n=b)

    2.

    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)]

    由指数的性质

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

    又因为指数函数是单调函数,所以

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

    3.与2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)

    4.与2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)

    其他性质:

    性质一:换底公式

    log(a)(N)=log(b)(N) / log(b)(a)

    推导如下

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

    a = b^[log(b)(a)]

    综合两式可得

    N = {b^[log(b)(a)]}^[log(a)(N)] = b^{[log(a)(N)]*[log(b)(a)]}

    又因为N=b^[log(b)(N)]

    所以

    b^[log(b)(N)] = b^{[log(a)(N)]*[log(b)(a)]}

    所以

    log(b)(N) = [log(a)(N)]*[log(b)(a)] {这步不明白或有疑问看上面的}

    所以log(a)(N)=log(b)(N) / log(b)(a)

    性质二:(不知道什么名字)

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

    推导如下

    由换底公式[lnx是log(e)(x),e称作自然对数的底]

    log(a^n)(b^m)=ln(a^n) / ln(b^n)

    由基本性质4可得

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

    再由换底公式

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

    --------------------------------------------(性质及推导 完 )

    公式三:

    log(a)(b)=1/log(b)(a)

    证明如下:

    由换底公式 log(a)(b)=log(b)(b)/log(b)(a) ----取以b为底的对数,log(b)(b)=1

    =1/log(b)(a)

    还可变形得:

    log(a)(b)*log(b)(a)=1