Solve 3^(2x^(2)-7x + 7) = 9

Explanation

To solve the equation \( 3^{2x^2 - 7x + 7} = 9 \), we can follow these steps:

### Step 1: Rewrite the equation
We know that \( 9 \) can be expressed as a power of \( 3 \):
\[ 9 = 3^2 \]
Thus, we can rewrite the equation as:
\[ 3^{2x^2 - 7x + 7} = 3^2 \]

### Step 2: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ 2x^2 - 7x + 7 = 2 \]

### Step 3: Rearrange the equation
Now, we will move \( 2 \) to the left side of the equation:
\[ 2x^2 - 7x + 7 - 2 = 0 \]
This simplifies to:
\[ 2x^2 - 7x + 5 = 0 \]

### Step 4: Factor the quadratic equation
Next, we will factor the quadratic equation \( 2x^2 - 7x + 5 = 0 \). We need to find two numbers that multiply to \( 2 \times 5 = 10 \) and add to \( -7 \). The numbers that satisfy this are \( -5 \) and \( -2 \).

Thus, we can factor it as:
\[ (2x - 5)(x - 1) = 0 \]

### Step 5: Solve for \( x \)
Now, we will set each factor equal to zero:
1. \( 2x - 5 = 0 \)
\[ 2x = 5 \implies x = \frac{5}{2} \]
2. \( x - 1 = 0 \)
\[ x = 1 \]

### Step 6: Write the final solutions
The solutions to the equation \( 3^{2x^2 - 7x + 7} = 9 \) are:
\[ x = 1 \quad \text{and} \quad x = \frac{5}{2} \]

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