To solve the equation \( 2^{2x} - 2^{x+2} = 12 \), let's proceed step by step.

### Step 1: Express the equation in terms of a single variable
Notice that both terms involve powers of 2. We can simplify this by setting a substitution:

Let \( y = 2^x \). This means:
- \( 2^{2x} = (2^x)^2 = y^2 \)
- \( 2^{x+2} = 2^x \cdot 2^2 = 4y \)

So the equation becomes:

\[
y^2 - 4y = 12
\]

### Step 2: Rearrange the equation into standard quadratic form
Rearrange the equation to get:

\[
y^2 - 4y - 12 = 0
\]

### Step 3: Solve the quadratic equation
To solve this quadratic equation, we can use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -4 \), and \( c = -12 \). Plugging these values into the formula:

\[
y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)}
\]

\[
y = \frac{4 \pm \sqrt{16 + 48}}{2}
\]

\[
y = \frac{4 \pm \sqrt{64}}{2}
\]

\[
y = \frac{4 \pm 8}{2}
\]

So we have two possible solutions for \( y \):

\[
y = \frac{4 + 8}{2} = 6 \quad \text{or} \quad y = \frac{4 - 8}{2} = -2
\]

### Step 4: Interpret the solutions
Since \( y = 2^x \), and \( 2^x \) is always positive, we discard \( y = -2 \) as a solution because \( 2^x \) cannot be negative.

So, \( y = 6 \) is the valid solution.

### Step 5: Solve for \( x \)
Now, recall \( y = 2^x \), so:

\[
2^x = 6
\]

To solve for \( x \), take the logarithm (base 2) of both sides:

\[
x = \log_2{6}
\]

Alternatively, you can use the change of base formula:

\[
x = \frac{\log{6}}{\log{2}}
\]

This gives the exact solution for \( x \). If you compute it:

\[
x \approx \frac{0.778}{0.301} \approx 2.29
\]

### Final Answer:
The solution is \( x \approx 2.29 \).

Would you like further details or have any questions?

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Here are 8 related questions to consider:
1. How would the solution change if the equation were \( 2^{2x} - 2^{x+2} = 0 \)?
2. What is the significance of the quadratic form in solving exponential equations?
3. How would you solve \( 3^{2x} - 3^{x+2} = 12 \) instead?
4. How can you verify the solution \( x \approx 2.29 \) by substitution?
5. What if the equation had a different base, such as \( 10^x \)?
6. How does logarithm base conversion affect the accuracy of the solution?
7. Can you solve the equation using a different method, like graphing?
8. What would happen if the right-hand side of the equation was negative?

**Tip:** When solving equations with exponents, substituting the exponential expression with a simpler variable can help transform the problem into a quadratic equation, which is often easier to solve.

Learn how to solve the exponential equation 2^(2x) - 2^(x+2) = 12 with detailed step-by-step instructions. This problem involves transforming exponential equations into quadratic form and applying logarithms to find the exact solution. Suitable for high school students in Grades 10-12.

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