Solve each of the following equatins :
`2^(2x+3)-57=65(2^(x)-1)`

Answer

To solve the equation \( 2^{2x+3} - 57 = 65(2^x - 1) \), we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the left side of the equation:
\[ 2^{2x+3} = 2^{2x} \cdot 2^3 = 8 \cdot 2^{2x} \]
So, the equation becomes:
\[ 8 \cdot 2^{2x} - 57 = 65(2^x - 1) \]

### Step 2: Substitute \( 2^x \)
Let \( d = 2^x \). Then, \( 2^{2x} = (2^x)^2 = d^2 \). Substituting this into the equation gives:
\[ 8d^2 - 57 = 65(d - 1) \]

### Step 3: Expand and rearrange
Expanding the right side:
\[ 8d^2 - 57 = 65d - 65 \]
Now, rearranging the equation to bring all terms to one side:
\[ 8d^2 - 65d - 57 + 65 = 0 \]
This simplifies to:
\[ 8d^2 - 65d + 8 = 0 \]

### Step 4: Factor or use the quadratic formula
To solve the quadratic equation \( 8d^2 - 65d + 8 = 0 \), we can use the quadratic formula:
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 8 \), \( b = -65 \), and \( c = 8 \).

### Step 5: Calculate the discriminant
First, calculate the discriminant:
\[ b^2 - 4ac = (-65)^2 - 4 \cdot 8 \cdot 8 = 4225 - 256 = 3969 \]

### Step 6: Solve for \( d \)
Now substitute back into the quadratic formula:
\[ d = \frac{65 \pm \sqrt{3969}}{2 \cdot 8} \]
Calculating \( \sqrt{3969} = 63 \):
\[ d = \frac{65 \pm 63}{16} \]
This gives us two possible solutions for \( d \):
1. \( d = \frac{128}{16} = 8 \)
2. \( d = \frac{2}{16} = \frac{1}{8} \)

### Step 7: Back substitute to find \( x \)
Now, recall that \( d = 2^x \):
1. For \( d = 8 \):
\[ 2^x = 8 \implies x = 3 \]
2. For \( d = \frac{1}{8} \):
\[ 2^x = \frac{1}{8} \implies 2^x = 2^{-3} \implies x = -3 \]

### Final Solution
The solutions to the equation are:
\[ x = 3 \quad \text{or} \quad x = -3 \]

---