(1) $$ 2^{2 x}=2^{2}\left(2^{x}+8\right) $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
2^{2x} = 2^{2}\left(2^{x} + 8\right)
$$

let's follow these steps:

### Step 1: Simplify the Equation

First, recognize that $$2^{2x}$$ can be rewritten as $$(2^{x})^{2}$$:

$$
(2^{x})^{2} = 2^{2} \left(2^{x} + 8\right)
$$

### Step 2: Substitute for Simplification

Let’s make a substitution to simplify the equation. Let:

$$
y = 2^{x}
$$

The equation becomes:

$$
y^{2} = 4(y + 8)
$$

### Step 3: Rearrange the Equation

Bring all terms to one side to form a quadratic equation:

$$
y^{2} - 4y - 32 = 0
$$

### Step 4: Solve the Quadratic Equation

Use the quadratic formula to solve for $$y$$:

$$
y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1}
$$

$$
y = \frac{4 \pm \sqrt{16 + 128}}{2}
$$

$$
y = \frac{4 \pm \sqrt{144}}{2}
$$

$$
y = \frac{4 \pm 12}{2}
$$

This gives two possible solutions:

1. $$ y = \frac{4 + 12}{2} = \frac{16}{2} = 8 $$
2. $$ y = \frac{4 - 12}{2} = \frac{-8}{2} = -4 $$

### Step 5: Consider the Validity of Solutions

Since $$ y = 2^{x} $$ and any exponential expression with base 2 is always positive, $$ y = -4 $$ is not a valid solution. Therefore, we have:

$$
y = 8
$$

### Step 6: Solve for $$ x $$

Recall that $$ y = 2^{x} $$:

$$
2^{x} = 8
$$

Since $$ 8 = 2^{3} $$, we have:

$$
2^{x} = 2^{3} \implies x = 3
$$

### Conclusion

The solution to the equation is:

$$
\boxed{3}
$$
The solution is $$ x = 3 $$.

(1) \( 2^{2 x}=2^{2}\left(2^{x}+8\right) \)