If f is a real function defined by `f(x)=(log(2x+1))/(sqrt3-x)`, then the domain of the function is

Options

A. `(-(1)/(2),oo)`
B. `(oo,3)`
C. `(-(1)/(2),3]`
D. `(-(1)/(2),3)`

Answer

To find the domain of the function \( f(x) = \frac{\log(2x + 1)}{\sqrt{3} - x} \), we need to consider the conditions under which the function is defined.

### Step 1: Determine the condition for the logarithm
The logarithm function \( \log(2x + 1) \) is defined only when its argument is positive. Therefore, we need:

\[ 2x + 1 > 0 \]

### Step 2: Solve the inequality
To solve the inequality \( 2x + 1 > 0 \):

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

### Step 3: Determine the condition for the denominator
Next, we need to ensure that the denominator \( \sqrt{3} - x \) is not equal to zero:

\[ \sqrt{3} - x \neq 0 \]

This implies:

\[ x \neq \sqrt{3} \]

### Step 4: Combine the conditions
Now, we combine the conditions from Steps 2 and 3. The function is defined for:

1. \( x > -\frac{1}{2} \)
2. \( x \neq \sqrt{3} \)

### Step 5: Write the domain in interval notation
The domain of the function can be expressed in interval notation as:

\[ \left(-\frac{1}{2}, \sqrt{3}\right) \cup \left(\sqrt{3}, \infty\right) \]

### Final Answer
Thus, the domain of the function \( f(x) = \frac{\log(2x + 1)}{\sqrt{3} - x} \) is:

\[ \left(-\frac{1}{2}, \sqrt{3}\right) \cup \left(\sqrt{3}, \infty\right) \]

---