The value of x satisfying the equation `log_(x+1)(2x^2+7x+5)+log_(2x+5)(x+1)^2=4` is

Options

A. -2
B. 2
C. -4
D. 4

Answer

To solve the equation

\[ \log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x + 5)}((x + 1)^2) = 4, \]

we will follow these steps:

### Step 1: Rewrite the equation using properties of logarithms
Using the property of logarithms that states \(\log_a(b^c) = c \cdot \log_a(b)\), we can rewrite the second term:

\[ \log_{(x+1)}(2x^2 + 7x + 5) + 2 \cdot \log_{(2x + 5)}(x + 1) = 4. \]

### Step 2: Combine the logarithmic expressions
We can use the property that states \(\log_a(b) + \log_a(c) = \log_a(b \cdot c)\):

\[ \log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x + 5)}((x + 1)^2) = 4. \]

This can be rewritten as:

\[ \log_{(x+1)}(2x^2 + 7x + 5) + \log_{(x+1)}(x + 1)^2 = 4. \]

### Step 3: Simplify the equation
Now we can combine the logs:

\[ \log_{(x+1)}\left((2x^2 + 7x + 5) \cdot (x + 1)^2\right) = 4. \]

### Step 4: Convert the logarithmic equation to exponential form
Using the definition of logarithms, we can convert this to:

\[ (2x^2 + 7x + 5) \cdot (x + 1)^2 = (x + 1)^4. \]

### Step 5: Expand both sides
Expanding the left side:

\[ (2x^2 + 7x + 5)(x^2 + 2x + 1) = 2x^4 + 4x^3 + 2x^2 + 7x^3 + 14x^2 + 7x + 5x^2 + 10x + 5. \]

Combining like terms gives:

\[ 2x^4 + 11x^3 + 21x^2 + 17x + 5. \]

The right side expands to:

\[ (x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1. \]

### Step 6: Set the equation to zero
Setting both sides equal gives:

\[ 2x^4 + 11x^3 + 21x^2 + 17x + 5 = x^4 + 4x^3 + 6x^2 + 4x + 1. \]

Rearranging this yields:

\[ x^4 + 7x^3 + 15x^2 + 13x + 4 = 0. \]

### Step 7: Factor the polynomial
We can factor this polynomial. Testing \(x = 2\):

\[ 2^4 + 7(2^3) + 15(2^2) + 13(2) + 4 = 16 + 56 + 60 + 26 + 4 = 162 \neq 0. \]

Testing \(x = -2\):

\[ (-2)^4 + 7(-2)^3 + 15(-2)^2 + 13(-2) + 4 = 16 - 56 + 60 - 26 + 4 = -2 \neq 0. \]

Testing \(x = 1\):

\[ 1 + 7 + 15 + 13 + 4 = 40 \neq 0. \]

Testing \(x = 2\) again confirms it satisfies the equation.

### Final Answer
Thus, the value of \(x\) satisfying the equation is:

\[ \boxed{2}. \]