lo g a m+lo g a (1+
1
m )+lo g a (1+
1
m+1 )+…+lo g a (1+
1
m+n-1 )
=log am+[log a (m+1)-log a m]+[log a (m+2)-log a (m+1)]+…+[log a (m+n)-log a (m+n-1)]
=log a (m+n),
则log a (m+n)=log am+log an=log a(mn),即m+n=mn.
故答案为:mn.
lo g a m+lo g a (1+
1
m )+lo g a (1+
1
m+1 )+…+lo g a (1+
1
m+n-1 )
=log am+[log a (m+1)-log a m]+[log a (m+2)-log a (m+1)]+…+[log a (m+n)-log a (m+n-1)]
=log a (m+n),
则log a (m+n)=log am+log an=log a(mn),即m+n=mn.
故答案为:mn.