The number of solution of the equation `log(-2x)= 2 log(x+1)` are

Options

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

Explanation

To solve the equation \( \log(-2x) = 2 \log(x+1) \), we will go through the following steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \log(-2x) = 2 \log(x+1) \]
Using the logarithmic identity \( a \log b = \log(b^a) \), we can rewrite the right side:
\[ \log(-2x) = \log((x+1)^2) \]

### Step 2: Set the arguments equal
Since the logarithmic function is one-to-one, we can set the arguments equal to each other:
\[ -2x = (x+1)^2 \]

### Step 3: Expand the right side
Now, we expand the right side:
\[ -2x = x^2 + 2x + 1 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 0 = x^2 + 2x + 1 + 2x \]
\[ 0 = x^2 + 4x + 1 \]

### Step 5: Apply the quadratic formula
Now we will apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 4, c = 1 \):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 - 4}}{2} \]
\[ x = \frac{-4 \pm \sqrt{12}}{2} \]
\[ x = \frac{-4 \pm 2\sqrt{3}}{2} \]
\[ x = -2 \pm \sqrt{3} \]

### Step 6: Identify the potential solutions
Thus, the potential solutions are:
\[ x = -2 + \sqrt{3} \quad \text{and} \quad x = -2 - \sqrt{3} \]

### Step 7: Check the validity of solutions
We need to check if these solutions are valid in the context of the logarithmic functions:
1. For \( \log(-2x) \) to be defined, \( -2x > 0 \) implies \( x < 0 \).
2. For \( \log(x+1) \) to be defined, \( x + 1 > 0 \) implies \( x > -1 \).

Now, we evaluate the solutions:
- \( x = -2 + \sqrt{3} \) (approximately \( -0.268 \)), which satisfies \( -1 < -2 + \sqrt{3} < 0 \).
- \( x = -2 - \sqrt{3} \) (approximately \( -3.732 \)), which does not satisfy \( x > -1 \).

### Conclusion
Only one solution \( x = -2 + \sqrt{3} \) is valid. Therefore, the number of solutions to the equation is:
\[ \boxed{1} \]