| Certified Educator
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)
| Certified Educator
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).
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