To solve the equation [tex]\( 4^x - 2^x - 12 = 0 \)[/tex], let's go through the problem step by step.
First, we can rewrite [tex]\( 4^x \)[/tex] in terms of a common base with [tex]\( 2^x \)[/tex]. Notice that [tex]\( 4 = 2^2 \)[/tex], so we have:
[tex]\[ 4^x = (2^2)^x = (2^x)^2 \][/tex]
Substituting this back into the equation, we get:
[tex]\[ (2^x)^2 - 2^x - 12 = 0 \][/tex]
Let [tex]\( y = 2^x \)[/tex]. Our equation then becomes a quadratic equation in terms of [tex]\( y \)[/tex]:
[tex]\[ y^2 - y - 12 = 0 \][/tex]
Now, we can solve this quadratic equation using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -12 \)[/tex].
Plugging in the values, we get:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-12)}}{2(1)} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{1 + 48}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{49}}{2} \][/tex]
[tex]\[ y = \frac{1 \pm 7}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1 + 7}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ y = \frac{1 - 7}{2} = \frac{-6}{2} = -3 \][/tex]
Since [tex]\( y = 2^x \)[/tex], we now substitute back to find [tex]\( x \)[/tex]:
1. For [tex]\( y = 4 \)[/tex]:
[tex]\[ 2^x = 4 \][/tex]
[tex]\[ 2^x = 2^2 \][/tex]
[tex]\[ x = 2 \][/tex]
2. For [tex]\( y = -3 \)[/tex]:
[tex]\[ 2^x = -3 \][/tex]
Since [tex]\( 2^x \)[/tex] is always positive for real [tex]\( x \)[/tex], there are no real solutions for this case. Thus, we disregard this solution.
So, the only real solution from our initial transformation is [tex]\( x = 2 \)[/tex].
### Checking Against Provided Possible Answers
The provided possible answers are: [tex]\( x = -2 \)[/tex], [tex]\( x = 1 \)[/tex], [tex]\( x = 5 \)[/tex], and [tex]\( x = 2 \)[/tex].
Among these, only [tex]\( x = 2 \)[/tex] is a valid solution.
### The Solution
Therefore, the correct answer for [tex]\( 4^x - 2^x - 12 = 0 \)[/tex] is [tex]\( x = 2 \)[/tex].
