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.