The domain of `log_(x^2 +x-2)[x+1/2 ]` is

Options

A. `(0,oo)`
B. `(1,oo)`
C. `(1,oo)`
D. `(1,oo)-{(-1 + sqrt(13))/(2)}`

Answer

To find the domain of the function \( \log_{(x^2 + x - 2)}\left(x + \frac{1}{2}\right) \), we need to ensure that the logarithmic function is defined. This requires satisfying three conditions:

1. The argument of the logarithm must be greater than 0.
2. The base of the logarithm must be greater than 0.
3. The base of the logarithm must not be equal to 1.

Let's solve these conditions step by step.

### Step 1: Argument of the logarithm must be greater than 0

The argument is \( x + \frac{1}{2} \). Therefore, we need:

\[ x + \frac{1}{2} > 0 \]

Solving this inequality:

\[ x > -\frac{1}{2} \]

### Step 2: Base of the logarithm must be greater than 0

The base is \( x^2 + x - 2 \). We need:

\[ x^2 + x - 2 > 0 \]

Factoring the quadratic expression:

\[ (x - 1)(x + 2) > 0 \]

To determine the intervals where this inequality holds, we find the roots by setting the expression equal to zero:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

Now, we test the intervals determined by these roots: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \).

- For \( x < -2 \) (e.g., \( x = -3 \)):
\((x - 1)(x + 2) = (-)(-) = +\) (True)

- For \( -2 < x < 1 \) (e.g., \( x = 0 \)):
\((x - 1)(x + 2) = (-)(+) = -\) (False)

- For \( x > 1 \) (e.g., \( x = 2 \)):
\((x - 1)(x + 2) = (+)(+) = +\) (True)

Thus, the solution for this inequality is:

\[ x \in (-\infty, -2) \cup (1, \infty) \]

### Step 3: Base of the logarithm must not be equal to 1

We need to ensure that:

\[ x^2 + x - 2 \neq 1 \]

This simplifies to:

\[ x^2 + x - 3 \neq 0 \]

Factoring gives:

\[ (x - \sqrt{4})(x + \sqrt{4}) \neq 0 \]

Finding the roots:

\[ x^2 + x - 3 = 0 \quad \Rightarrow \quad x = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2} \]

Thus, the values that \( x \) cannot take are:

\[ x \neq \frac{-1 + \sqrt{13}}{2} \quad \text{and} \quad x \neq \frac{-1 - \sqrt{13}}{2} \]

### Step 4: Combine all conditions

Now we need to find the intersection of the three conditions:

1. From Step 1: \( x > -\frac{1}{2} \)
2. From Step 2: \( x \in (-\infty, -2) \cup (1, \infty) \)
3. From Step 3: Exclude \( \frac{-1 + \sqrt{13}}{2} \) and \( \frac{-1 - \sqrt{13}}{2} \)

The only interval that satisfies all conditions is:

\[ (1, \infty) \]

Thus, the domain of the function \( \log_{(x^2 + x - 2)}\left(x + \frac{1}{2}\right) \) is:

\[ \boxed{(1, \infty)} \]