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);

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)