If `log_(1/2) ((x^2+6x+9)/(2(x+1)) )< - log_2(x+1)` then complete set of values of x is

Options

A. `(-1,1+2sqrt(2))`
B. `(1-2sqrt(2),2)`
C. `(-1,oo)`
D. `(1-2sqrt(2),1+2sqrt(2))`

Answer

To solve the inequality \( \log_{1/2} \left( \frac{x^2 + 6x + 9}{2(x+1)} \right) < -\log_2(x+1) \), we will follow these steps:

### Step 1: Rewrite the logarithm
Using the property of logarithms, we can rewrite the left-hand side:
\[ \log_{1/2} \left( \frac{x^2 + 6x + 9}{2(x+1)} \right) = -\log_2 \left( \frac{x^2 + 6x + 9}{2(x+1)} \right) \]
Thus, the inequality becomes:
\[ -\log_2 \left( \frac{x^2 + 6x + 9}{2(x+1)} \right) < -\log_2(x+1) \]

### Step 2: Remove the negative signs
Multiplying both sides by -1 reverses the inequality:
\[ \log_2 \left( \frac{x^2 + 6x + 9}{2(x+1)} \right) > \log_2(x+1) \]

### Step 3: Exponentiate both sides
Since the logarithm is a monotonically increasing function, we can exponentiate both sides:
\[ \frac{x^2 + 6x + 9}{2(x+1)} > x + 1 \]

### Step 4: Simplify the inequality
Cross-multiplying gives:
\[ x^2 + 6x + 9 > 2(x + 1)(x + 1) \]
Expanding the right-hand side:
\[ x^2 + 6x + 9 > 2(x^2 + 2x + 1) \]
\[ x^2 + 6x + 9 > 2x^2 + 4x + 2 \]

### Step 5: Rearranging the terms
Rearranging the inequality:
\[ 0 > 2x^2 + 4x + 2 - x^2 - 6x - 9 \]
\[ 0 > x^2 - 2x - 7 \]
This can be rewritten as:
\[ x^2 - 2x - 7 < 0 \]

### Step 6: Solve the quadratic inequality
To find the roots of the equation \( x^2 - 2x - 7 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 + 28}}{2} = \frac{2 \pm \sqrt{32}}{2} = \frac{2 \pm 4\sqrt{2}}{2} = 1 \pm 2\sqrt{2} \]
Thus, the roots are \( x_1 = 1 - 2\sqrt{2} \) and \( x_2 = 1 + 2\sqrt{2} \).

### Step 7: Determine the intervals
The quadratic \( x^2 - 2x - 7 \) opens upwards (since the coefficient of \( x^2 \) is positive). Therefore, it is negative between the roots:
\[ 1 - 2\sqrt{2} < x < 1 + 2\sqrt{2} \]

### Step 8: Consider the domain of the logarithm
We must also consider the condition for the logarithm to be defined:
\[ x + 1 > 0 \implies x > -1 \]

### Step 9: Find the intersection of intervals
Now we need to find the intersection of \( (-1, 1 + 2\sqrt{2}) \) and \( (1 - 2\sqrt{2}, 1 + 2\sqrt{2}) \):
- \( 1 - 2\sqrt{2} \) is approximately \( -1.828 \), which is less than \(-1\).
- Therefore, the intersection is:
\[ (-1, 1 + 2\sqrt{2}) \]

### Final Solution
The complete set of values of \( x \) that satisfy the inequality is:
\[ x \in (-1, 1 + 2\sqrt{2}) \]