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 follow these steps:

### Step 1: Rewrite the equation using logarithmic properties
Using the property of logarithms that states \( a \log b = \log(b^a) \), we can rewrite the right side of the equation:
\[ \log(-2x) = \log((x+1)^2) \]

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

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

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

### Step 5: Solve the quadratic equation
We will use 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: Determine the approximate values of the roots
Calculating the approximate values:
1. \( x_1 = -2 + \sqrt{3} \approx -2 + 1.732 \approx -0.268 \)
2. \( x_2 = -2 - \sqrt{3} \approx -2 - 1.732 \approx -3.732 \)

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

Thus, the valid range for \( x \) is:
\[ -1 < x < 0 \]

### Step 8: Analyze the roots
- \( x_1 \approx -0.268 \) is in the valid range.
- \( x_2 \approx -3.732 \) is not in the valid range.

### Conclusion
There is only one valid solution for the equation \( \log(-2x) = 2 \log(x+1) \).

### Final Answer
The number of solutions is **1**.

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