Correct Option: B

If x(n) ←z→ X(z)

then nx(n) ←z→ – z	d	X(z)	az– 1
	dz		1 + az– 1

and if X (Z) = log (1 + az^{– 1})

then nx(n) ←z→ – z	d	X(z)	az– 1
	dZ		1 + az– 1

As we know that

– an u(n) ←z→	1
	1 + az– 1

Similarly, by using time-shifting property.

(– a)n– 1 u(n – 1) ←z→	1	· z– 1
	1 + az– 1

or a(– a)n– 1 u(n – 1) =	az– 1
	1 + az– 1

or – (– a)n u(n – 1) =	az– 1
	1 + az– 1

or – (– a)ⁿ u(n – 1) = nx(n)

or x(n) =	– (– a)n	u(n – 1)
	n

Hence, alternative (B) is the correct choice.