Find the values of x satisfying `log_(x^(2)+6x+8)log_(2x^(2)+2x+3)(x^(2)-2x)=0` is

Options

A. A)0
B. B)`-1`
C. C)2
D. D)`-3`

Answer

To solve the equation \( \log_{(x^2 + 6x + 8)} \left( \log_{(2x^2 + 2x + 3)} (x^2 - 2x) \right) = 0 \), we will follow these steps:

### Step 1: Understand the logarithmic equation
The equation \( \log_{a}(b) = 0 \) implies that \( b = 1 \) (since any logarithm equals zero when its argument is 1). Therefore, we can rewrite our equation as:
\[ \log_{(2x^2 + 2x + 3)} (x^2 - 2x) = 1 \]

### Step 2: Set the inner logarithm equal to 1
From the previous step, we have:
\[ x^2 - 2x = 2x^2 + 2x + 3 \]

### Step 3: Rearrange the equation
Now, we will rearrange the equation:
\[ x^2 - 2x - 2x^2 - 2x - 3 = 0 \]
This simplifies to:
\[ -x^2 - 4x - 3 = 0 \]
Multiplying through by -1 gives:
\[ x^2 + 4x + 3 = 0 \]

### Step 4: Factor the quadratic equation
Next, we will factor the quadratic equation:
\[ x^2 + 4x + 3 = (x + 3)(x + 1) = 0 \]

### Step 5: Solve for x
Setting each factor to zero gives us:
1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x + 1 = 0 \) → \( x = -1 \)

### Step 6: Verify the solutions
We need to ensure that the bases of the logarithms are valid (i.e., positive and not equal to 1):
- For \( x = -3 \):
- \( x^2 + 6x + 8 = 9 - 18 + 8 = -1 \) (not valid)
- \( 2x^2 + 2x + 3 = 18 - 6 + 3 = 15 \) (valid)

- For \( x = -1 \):
- \( x^2 + 6x + 8 = 1 - 6 + 8 = 3 \) (valid)
- \( 2x^2 + 2x + 3 = 2 - 2 + 3 = 3 \) (valid)

### Conclusion
The only valid solution is:
\[ x = -1 \]