Find domain of the function
`f(x) = sqrt(-log_((x+4)/2)(log_2(2x-1)/(3+x))`

Options

A. `(-4, -3) uu (4 , oo)`
B. `(-oo, -3) uu (4,oo)`
C. `(-oo,-4)uu(3,oo)`
D. Noen of these

Answer

To find the domain of the function

\[ f(x) = \sqrt{-\log_{\frac{x+4}{2}}\left(\frac{\log_2(2x-1)}{3+x}\right)}, \]

we need to ensure that the expression inside the square root is non-negative and that the logarithmic expressions are defined. Let's break this down step by step.

### Step 1: Ensure the expression inside the square root is non-negative

The expression inside the square root must be greater than or equal to zero:

\[ -\log_{\frac{x+4}{2}}\left(\frac{\log_2(2x-1)}{3+x}\right) \geq 0. \]

This implies:

\[ \log_{\frac{x+4}{2}}\left(\frac{\log_2(2x-1)}{3+x}\right} \leq 0. \]

### Step 2: Analyze the logarithmic condition

The logarithm is less than or equal to zero when the argument is between 0 and 1. Thus, we need:

1. \(\frac{\log_2(2x-1)}{3+x} < 1\)
2. \(\frac{\log_2(2x-1)}{3+x} > 0\)

### Step 3: Solve the first inequality

Starting with the first inequality:

\[ \frac{\log_2(2x-1)}{3+x} < 1 \implies \log_2(2x-1) < 3+x. \]

This can be rearranged to:

\[ 2x - 1 < 2^{3+x}. \]

### Step 4: Solve the second inequality

Now for the second inequality:

\[ \frac{\log_2(2x-1)}{3+x} > 0 \implies \log_2(2x-1) > 0. \]

This means:

\[ 2x - 1 > 1 \implies 2x > 2 \implies x > 1. \]

### Step 5: Ensure the logarithm base is valid

The base of the logarithm, \(\frac{x+4}{2}\), must be greater than 0 and not equal to 1:

1. \(\frac{x+4}{2} > 0 \implies x + 4 > 0 \implies x > -4\).
2. \(\frac{x+4}{2} \neq 1 \implies x + 4 \neq 2 \implies x \neq -2\).

### Step 6: Combine the conditions

From the inequalities, we have:

- \(x > 1\) (from the second inequality).
- \(x > -4\) (which is always satisfied if \(x > 1\)).
- \(x \neq -2\) (which is also satisfied since \(-2 < 1\)).

### Step 7: Final domain

Thus, the only relevant condition is \(x > 1\).

The domain of the function \(f(x)\) is:

\[ \text{Domain: } (1, \infty). \]