求证logaM/N=logaM-logaN

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求证logaM/N=logaM-logaN求证logaM/N=logaM-logaN求证logaM/N=logaM-logaN设a^x=M,a^y=NM/N=(a^x)/(a^y)=a^(x-y)lo

求证logaM/N=logaM-logaN
求证logaM/N=logaM-logaN

求证logaM/N=logaM-logaN
设a^x=M,a^y=N
M/N=(a^x)/(a^y)=a^(x-y)
loga (M/N)=loga a^(x-y)=x-y
loga M-loga N=x-y
所以loga (M/N)=loga M-loga N

=loga[M^logaN] =logaN*logaM =logaM*logaN =loga[N^logaM] 所以x=N^logaM 所以M^logaN=N^logaM M^logaN=N^logaM