2^(2x) + 2^(x+2) - 12 = 0?

Question

Precalculus problem
logarithmiac/exponential equation
a step-by step solution please

Answer 1

2^(2x) + 2^(x+2) - 12 = 0

First, use the laws of exponents to write this as a function of 2^x:

(2^x)² + 4*2^x - 12 = 0

Complete the square by adding 16 to both sides:

(2^x)² + 4*2^x + 4 = 16

Factor:

(2^x + 2)² = 16

Take the square roots of both sides:

2^x + 2 = ±4

Subtract 2:

2^x = -2±4

So 2^x = 2 or 2^x = -6. However, exponentials are never negative, so the second equation is impossible. We therefore conclude that:

2^x = 2
x=1

And we are done.

Answer 2

2^(2x) + 2^(x+2) - 12 = (2^x + 2)^2 - 16 = 0

So, 2^x + 2 = +/- 4, 2^x = 2, -6 and x = 1.

Answer 3

2^(2x) + 2^(x+2) - 12 = 0
(2^x)^2 + (2^x)(2^2) - 12 = 0
(2^x)^2 + 4(2^x) - 12 = 0

Let y=2^x
y^2 + 4y - 12 = 0
(y+6)(y-2) = 0
y = -6 or 2
2^x = -6 or 2^x = 2

x = log2(-6) (NA), x = log2(2) => x = 1