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证明对数运算法则(1)log(a)(MN)=log(a)(M)+log(a)(N);   (2)log(a)(M/N)=log(a)(M)-log(a)(N);
(1)log(a)(MN)=log(a)(M)+log(a)(N);   (2)log(a)(M/N)=log(a)(M)-log(a)(N);   (3)log(a)(M^n)=nlog(a)(M) (n∈R)
人气:474 ℃ 时间:2019-10-29 11:56:28
解答
证明:
设loga (M)=m,loga (N)=p
则 a^m=M ,a^n=p
(1) MN=a^m* a^p=a^(m+p)
所以 m+p=loga(MN)
即 log(a)(MN)=log(a)(M)+log(a)(N)
(2) M/N=a^m / a^p=a^(m-p)
所以 m-p=loga(M/N)
即 log(a)(M/N)=log(a)(M)-log(a)(N)
(3) M^n=(a^m)^n=a^(mn)
mn=loga (m^n)
即 log(a)(M^n)=nlog(a)(M)
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