X(z)=log(1−2z−1) |z|>2 dX(z)dz=2z−21−2z−1
−zdX(z)dz=−2z−11−2z−1
From differentiation property,
z[nx(n)]=−zdX(z)dz
nx[n]=z−1[−zdX(z)dz]=z−1[−2z−11−2z−1]
∴nx[n]=−2[2n−1u[n−1]]
∴x[n]=−2nu[n−1]n
∴x[2]=−222=−2