由f^-1=log2(1+x)/(1-x)可得
2^x=(1+y)/(1-y)
化簡得
f(x)=(2^x-1)/(2^x+1)
1-(2^x-1)/(2^x+1)>1/(4^x-1)
等價(jià)于2/(2^x+1)>1/(2^x+1)(2^x-1)
2^x+1>0,則(2^x-3/2)/(2^x-1)>0
即2^x>3/2或0小于2^x小于1
則解集為{x/x>log2(3/2)或x小于0}