Solve :
`8 xx 2^(2x) + 4 xx 2^(x +1) = 1 + 2^(x)`

Answer

To solve the equation \( 8 \cdot 2^{2x} + 4 \cdot 2^{x + 1} = 1 + 2^{x} \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 8 \cdot 2^{2x} + 4 \cdot 2^{x + 1} = 1 + 2^{x} \]

### Step 2: Simplify the terms
Notice that \( 4 \cdot 2^{x + 1} \) can be rewritten as \( 4 \cdot 2 \cdot 2^{x} = 8 \cdot 2^{x} \). Thus, we can rewrite the equation as:
\[ 8 \cdot 2^{2x} + 8 \cdot 2^{x} = 1 + 2^{x} \]

### Step 3: Move all terms to one side
Rearranging the equation gives:
\[ 8 \cdot 2^{2x} + 8 \cdot 2^{x} - 2^{x} - 1 = 0 \]
This simplifies to:
\[ 8 \cdot 2^{2x} + 7 \cdot 2^{x} - 1 = 0 \]

### Step 4: Substitute \( t = 2^{x} \)
Let \( t = 2^{x} \). Then, \( 2^{2x} = t^2 \). Substitute this into the equation:
\[ 8t^2 + 7t - 1 = 0 \]

### Step 5: Factor the quadratic equation
Now we need to factor the quadratic equation \( 8t^2 + 7t - 1 = 0 \). We can use the method of splitting the middle term:
\[ 8t^2 + 8t - t - 1 = 0 \]
Grouping gives:
\[ (8t^2 + 8t) + (-t - 1) = 0 \]
Factoring out common terms:
\[ 8t(t + 1) - 1(t + 1) = 0 \]
This can be factored as:
\[ (8t - 1)(t + 1) = 0 \]

### Step 6: Solve for \( t \)
Setting each factor to zero gives:
1. \( 8t - 1 = 0 \) → \( t = \frac{1}{8} \)
2. \( t + 1 = 0 \) → \( t = -1 \)

### Step 7: Back substitute for \( x \)
Recall that \( t = 2^{x} \):
1. From \( t = \frac{1}{8} \):
\[ 2^{x} = \frac{1}{8} \implies 2^{x} = 2^{-3} \implies x = -3 \]
2. From \( t = -1 \):
\[ 2^{x} = -1 \text{ (not possible, since the base of an exponent cannot be negative)} \]

### Final Solution
Thus, the only valid solution is:
\[ \boxed{-3} \]