Given the inequality:

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

First we will write: 16 = 2^4.

==> 2^4^(x-1) > 2^2x+2

We know that x^a^b = x^ab

==> 2^4(x-1) > 2^(2x+2)

==> 2^(4x-4) > 2^(2x+2)

Now that the bases are equal, then the powers should the inequality

==> 4x -4 >2x + 2

We will combine like terms.

==> 4x-2x > 2 + 4

==> 2x > 6

Divide by 2.

==> x > 3

Then the answer is:

**x belongs to the interval ( 3, inf)**

We have to solve the inequality 16^(x-1)>2^(2x+2)

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

=> 2^4*(x - 1) > 2^(2x + 2)

as the base is the same and positive we can write

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

=> 4x - 4 > 2x +2

=> 4x - 2x > 2 + 4

=> 2x > 6

=> x > 6/2

=> x > 3

**Therefore x > 3.**

We'll write both bases as power of 2:

2^4(x-1)>2^(2x+2)

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

4(x-1)> 2x+2

We'll divide by 2:

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 interval of values of x, for the inequality holds, is (3 , +infinite).**