If the equation ln `(x^(2) +5x ) -ln (x+a +3)=0` has exactly one solution for x, then possible integral value of a is:

Options

A. 4
B. 5
C. 6
D. 7

Answer

To solve the equation \( \ln(x^2 + 5x) - \ln(x + a + 3) = 0 \) for the condition that it has exactly one solution for \( x \), we can follow these steps:

### Step 1: Simplify the Equation
We start with the equation:
\[ \ln(x^2 + 5x) - \ln(x + a + 3) = 0 \]
Using the property of logarithms, we can combine the logs:
\[ \ln\left(\frac{x^2 + 5x}{x + a + 3}\right) = 0 \]
This implies:
\[ \frac{x^2 + 5x}{x + a + 3} = 1 \]

### Step 2: Remove the Logarithm
Cross-multiplying gives us:
\[ x^2 + 5x = x + a + 3 \]

### Step 3: Rearrange the Equation
Rearranging the equation leads to:
\[ x^2 + 5x - x - a - 3 = 0 \]
This simplifies to:
\[ x^2 + 4x - (a + 3) = 0 \]

### Step 4: Determine the Condition for One Solution
For the quadratic equation \( x^2 + 4x - (a + 3) = 0 \) to have exactly one solution, the discriminant must be zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 4 \), and \( c = -(a + 3) \). Therefore, we have:
\[ D = 4^2 - 4 \cdot 1 \cdot (-(a + 3)) = 16 + 4(a + 3) \]
Setting the discriminant to zero for exactly one solution:
\[ 16 + 4(a + 3) = 0 \]

### Step 5: Solve for \( a \)
Solving the equation:
\[ 16 + 4a + 12 = 0 \]
\[ 4a + 28 = 0 \]
\[ 4a = -28 \]
\[ a = -7 \]

### Step 6: Check the Validity of \( a \)
We need to ensure that the logarithmic expressions are defined. For \( \ln(x^2 + 5x) \) and \( \ln(x + a + 3) \) to be defined, we require:
1. \( x^2 + 5x > 0 \)
2. \( x + a + 3 > 0 \)

For \( a = -7 \):
\[ x + (-7) + 3 > 0 \Rightarrow x - 4 > 0 \Rightarrow x > 4 \]
The quadratic \( x^2 + 5x > 0 \) has roots at \( x = 0 \) and \( x = -5 \), thus it is positive for \( x < -5 \) and \( x > 0 \).

### Conclusion
The only integral value of \( a \) that satisfies the condition for exactly one solution is:
\[ \boxed{-7} \]