f(2^an)=2n
即 x=2^an
则log2(2^an)=an
logx(2)=log2(2)/log2(x)=1/an
==> an - 1/an=2n
==> an^2-2n* an -1=0
==> an=n+(n^2+1)^(1/2) an =n -(n^2+1)^(1/2)
嘿嘿,好象有点怪