Solve : `2^(2x +3) - 9 xx 2^(x) + 1 = 0`

Answer

To solve the equation \(2^{2x + 3} - 9 \cdot 2^x + 1 = 0\), we can follow these steps:

### Step 1: Rewrite the equation using properties of exponents
We can express \(2^{2x + 3}\) as \(2^{2x} \cdot 2^3\):
\[ 2^{2x + 3} = 2^{2x} \cdot 8 \]
Thus, the equation becomes:
\[ 8 \cdot 2^{2x} - 9 \cdot 2^x + 1 = 0 \]

### Step 2: Substitute \(2^x\) with a new variable
Let \(t = 2^x\). Then, \(2^{2x} = (2^x)^2 = t^2\). Substituting these into the equation gives:
\[ 8t^2 - 9t + 1 = 0 \]

### Step 3: Solve the quadratic equation
Now we have a quadratic equation in \(t\):
\[ 8t^2 - 9t + 1 = 0 \]
We can use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 8\), \(b = -9\), and \(c = 1\).

Calculating the discriminant:
\[ b^2 - 4ac = (-9)^2 - 4 \cdot 8 \cdot 1 = 81 - 32 = 49 \]

Now applying the quadratic formula:
\[ t = \frac{9 \pm \sqrt{49}}{2 \cdot 8} = \frac{9 \pm 7}{16} \]

Calculating the two possible values for \(t\):
1. \(t = \frac{9 + 7}{16} = \frac{16}{16} = 1\)
2. \(t = \frac{9 - 7}{16} = \frac{2}{16} = \frac{1}{8}\)

### Step 4: Substitute back to find \(x\)
Recall that \(t = 2^x\):
1. For \(t = 1\):
\[ 2^x = 1 \implies x = 0 \quad (\text{since } 2^0 = 1) \]
2. For \(t = \frac{1}{8}\):
\[ 2^x = \frac{1}{8} \implies 2^x = 2^{-3} \implies x = -3 \]

### Final Solution
The solutions to the equation are:
\[ x = 0 \quad \text{or} \quad x = -3 \]

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