The sum of all value of x so that `16^((x^(2)+_3x-1))=8^((x^(2)+3x+2))`, is

Options

A. 2
B. 3
C. `-3`
D. `-5`

Answer

To solve the equation \( 16^{(x^2 + 3x - 1)} = 8^{(x^2 + 3x + 2)} \), we can follow these steps:

### Step 1: Rewrite the bases in terms of powers of 2
We know that:
- \( 16 = 2^4 \)
- \( 8 = 2^3 \)

Thus, we can rewrite the equation as:
\[ (2^4)^{(x^2 + 3x - 1)} = (2^3)^{(x^2 + 3x + 2)} \]
This simplifies to:
\[ 2^{4(x^2 + 3x - 1)} = 2^{3(x^2 + 3x + 2)} \]

### Step 2: Set the exponents equal to each other
Since the bases are the same, we can equate the exponents:
\[ 4(x^2 + 3x - 1) = 3(x^2 + 3x + 2) \]

### Step 3: Expand both sides
Expanding both sides gives:
\[ 4x^2 + 12x - 4 = 3x^2 + 9x + 6 \]

### Step 4: Move all terms to one side
Rearranging the equation:
\[ 4x^2 - 3x^2 + 12x - 9x - 4 - 6 = 0 \]
This simplifies to:
\[ x^2 + 3x - 10 = 0 \]

### Step 5: Factor the quadratic equation
Now we need to factor the quadratic:
\[ x^2 + 5x - 2x - 10 = 0 \]
This can be factored as:
\[ (x + 5)(x - 2) = 0 \]

### Step 6: Solve for x
Setting each factor to zero gives us the solutions:
1. \( x + 5 = 0 \) → \( x = -5 \)
2. \( x - 2 = 0 \) → \( x = 2 \)

### Step 7: Find the sum of the solutions
Now, we need to find the sum of the solutions:
\[ x_1 + x_2 = -5 + 2 = -3 \]

### Final Answer
The sum of all values of \( x \) is:
\[ \boxed{-3} \]