Solve the equation:
`2^(2x+3)+2^(x+3)=1+2^(x)`.

Answer

To solve the equation \( 2^{2x+3} + 2^{x+3} = 1 + 2^x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2^{2x+3} + 2^{x+3} = 1 + 2^x \]

### Step 2: Simplify the exponents
Using the property of exponents, we can rewrite \( 2^{2x+3} \) and \( 2^{x+3} \):
\[ 2^{2x+3} = 2^{2x} \cdot 2^3 = 8 \cdot 2^{2x} \]
\[ 2^{x+3} = 2^x \cdot 2^3 = 8 \cdot 2^x \]
Substituting these back into the equation gives:
\[ 8 \cdot 2^{2x} + 8 \cdot 2^x = 1 + 2^x \]

### Step 3: Factor out common terms
Now, we can factor out \( 8 \) from the left side:
\[ 8(2^{2x} + 2^x) = 1 + 2^x \]

### Step 4: Rearrange the equation
Next, we can rearrange the equation to isolate terms:
\[ 8(2^{2x} + 2^x) - 2^x - 1 = 0 \]
This simplifies to:
\[ 8 \cdot 2^{2x} + 8 \cdot 2^x - 2^x - 1 = 0 \]
\[ 8 \cdot 2^{2x} + 7 \cdot 2^x - 1 = 0 \]

### Step 5: Substitute \( t = 2^x \)
Let \( t = 2^x \). Then, \( 2^{2x} = t^2 \). Substituting gives:
\[ 8t^2 + 7t - 1 = 0 \]

### Step 6: Solve the quadratic equation
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 8, b = 7, c = -1 \):
\[ t = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
Calculating the discriminant:
\[ 7^2 - 4 \cdot 8 \cdot (-1) = 49 + 32 = 81 \]
Thus,
\[ t = \frac{-7 \pm 9}{16} \]

### Step 7: Find the values of \( t \)
Calculating the two possible values:
1. \( t = \frac{2}{16} = \frac{1}{8} \)
2. \( t = \frac{-16}{16} = -1 \)

### Step 8: Solve for \( x \)
Since \( t = 2^x \), we can solve for \( x \):
1. For \( t = \frac{1}{8} \):
\[ 2^x = \frac{1}{8} \implies 2^x = 2^{-3} \implies x = -3 \]
2. For \( t = -1 \):
This is not possible since \( 2^x \) is always positive.

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