急！帮我证明一下这三个结论，看质量再重酬

log (a) (M/N) =log(a)M-log(a)N
假设M=a^x,N=a^y,则log (a) (M/N)=log(a)((a)^(x-y))=x-y=log(a)M-log(a)N
log(a)M＾n=n•log(a)M(n∈R)
假设M=a^x,则log(a)M＾n=log(a)(a^(nx))=nx=n•log(a)M(n∈R)

log(a＾u)M＾u=(v/u)log(a)M

假设M=a^x,则log(a＾v)M＾u=log(a＾v)(a^(ux))=log(a＾v)((a^v)^(ux/v))=ux/v

=(v/u)*x=(v/u)log(a)M

1.
要证：log (a) (M/N) =log(a)M-log(a)N
即证：a^[log (a) (M/N)]=a^[log(a)M-log(a)N]
M/N=a^[log(a)M]/a^[log(a)N]=M/N
得证

2.
log(a)M^n=nlog(a)M
<-
a^[log(a)M^n]=a^[nlog(a)M]
M^n=[a^log(a)M]^n=M^n
得证

3.
log(a^u)M^v=(v/u)log(a)M
log(a^u)M^u
=lnM^v/lna^u
=vlnM/(ulna)
=(v/u)log(a)M

M的次方应该是v吧
对数运算转化成指数运算就好理解了

1.
要证：log (a) (M/N) =log(a)M-log(a)N
即证：a^[log (a) (M/N)]=a^[log(a)M-log(a)N]
M/N=a^[log(a)M]/a^[log(a)N]=M/N
得证

2.
log(a)M^n=nlog(a)M
<-
a^[log(a)M^n]