We have to solve: 81^(x-1) - 9^(x+1) > 0

81^(x-1) - 9^(x+1) > 0

=> 9^2^(x - 1) - 9^(x + 1) > 0

=> 9^(2x - 2) - 9^(x + 1) > 0

=> 9^(2x - 2) > 9^(x + 1)

As 9 is a positive base

=> 2x - 2 > x + 1

=> x > 3

**The solution is x can take values that lie in (3 , +inf.)**

81^(x-1) -9^(x+1) > 0.

=> 81^(x-1) > 9^(x+1)

9^2(x-1) > 9^(x+1)

We compare the exponents as the bases are same .

2(x-1) > x+1

2x-2 > x+1

2x-x > 3

So **x> 3.**

We'll the base of the 1st term as power of 9:

9^2*(x-1) - 9^(x+1) > 0

We'll shift to the left, the 2nd term:

9^2*(x-1) > 9^(x+1)

Since the bases are bigger than unit value, the function is increasing and we'll get:

2*(x-1)> (x+1)

We'll remove the brackets:

2x - 2 > x + 1

We'll subtract x both sides:

2x - x - 2 > 1

x - 2 > 1

We'll add 2 both sides:

x > 2 + 1

x > 3

**The range of values of x, for the inequality to hold, is (3 , +infinite).**