Find the domain `f(x)=(log_(2x)3)/(cos^(- 1)(2x-1)`

Answer

To find the domain of the function \( f(x) = \frac{\log_{2x} 3}{\cos^{-1}(2x - 1)} \), we need to analyze the conditions for both the numerator and the denominator separately.

### Step 1: Analyze the numerator \( \log_{2x} 3 \)

The logarithmic function \( \log_{a} b \) is defined when:
1. The base \( a > 0 \) and \( a \neq 1 \).
2. The argument \( b > 0 \).

For our case, the base is \( 2x \) and the argument is \( 3 \).

#### Condition 1: Base \( 2x > 0 \)
\[ 2x > 0 \implies x > 0 \]

#### Condition 2: Base \( 2x \neq 1 \)
\[ 2x \neq 1 \implies x \neq \frac{1}{2} \]

So, from the numerator, we have:
- \( x > 0 \)
- \( x \neq \frac{1}{2} \)

### Step 2: Analyze the denominator \( \cos^{-1}(2x - 1) \)

The function \( \cos^{-1}(y) \) is defined for \( -1 \leq y \leq 1 \). Therefore, we need:
\[ -1 \leq 2x - 1 \leq 1 \]

#### Breaking this down into two inequalities:

1. \( 2x - 1 \geq -1 \)
\[ 2x \geq 0 \implies x \geq 0 \]

2. \( 2x - 1 \leq 1 \)
\[ 2x \leq 2 \implies x \leq 1 \]

So, from the denominator, we have:
- \( 0 \leq x \leq 1 \)

### Step 3: Combine the conditions

Now we combine the conditions from both the numerator and the denominator:
1. From the numerator: \( x > 0 \) and \( x \neq \frac{1}{2} \)
2. From the denominator: \( 0 \leq x \leq 1 \)

Thus, the combined conditions give us:
- \( 0 < x \leq 1 \) and \( x \neq \frac{1}{2} \)

### Step 4: Write the domain in interval notation

The domain of \( f(x) \) can be expressed in interval notation as:
\[ \text{Domain} = (0, \frac{1}{2}) \cup (\frac{1}{2}, 1] \]

### Final Answer:
The domain of the function \( f(x) = \frac{\log_{2x} 3}{\cos^{-1}(2x - 1)} \) is:
\[ (0, \frac{1}{2}) \cup (\frac{1}{2}, 1] \]

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