求证 log(a) (M·N)=log(a) M+log(a) N

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求证 log(a) (M·N)=log(a) M+log(a) N
求证 log(a) (M·N)=log(a) M+log(a) N

求证 log(a) (M·N)=log(a) M+log(a) N
证明:设log(a)(mn)=k 则有:a^k=mn
log(a)m=x 则有:a^x=m
log(a)n=y 则有:a^y=n
可得:mn=a^x*a^y=a^(x+y)=a^k
即:k=x+y
得:log(a)(mn)=log(a)m+log(a)n

设a的A次方为M,a的B次方为N,则MN=a的A+B次方,都取对数,log(a)(MN)=A+B=log(a)M+log(a)N 望采纳谢谢

log(a) (M·N)=log(a) M+log(a) N
a^x=M, ,,,,,,(1) a^y=N, ..........(2)
把(1)(2)化成对数式 x= log(a) ,y=Mlog(a) N
(1)*(2)得,a^x ·a^y=M·N
化成对数式得
a^(x+y)=M·N
log(a) (M·N)=x+y=log(a) M+log(a) N

设 M=a^p, N=a^q
log(a) (M·N)=log(a) a^(p+q)=p+q
log(a) M+log(a) N=p+q
∴ log(a) (M·N)=log(a) M+log(a) N