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

Explanation

To find the domain of the function \( f(x) = \sqrt{-\log_{\frac{x+4}{2}} \left( \log_2 \left( \frac{2x-1}{3+x} \right) \right)} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

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

This leads us to the condition:

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

### Step 1: Determine when the logarithm is defined

For the logarithm \( \log_2 \left( \frac{2x-1}{3+x} \right) \) to be defined, the argument must be positive:

\[ \frac{2x-1}{3+x} > 0 \]

This gives us two conditions:
1. \( 2x - 1 > 0 \) which simplifies to \( x > \frac{1}{2} \)
2. \( 3 + x > 0 \) which simplifies to \( x > -3 \)

Since \( x > \frac{1}{2} \) is the stricter condition, we have:

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

### Step 2: Analyze the logarithm's base

Next, we need to ensure that the base of the logarithm \( \frac{x+4}{2} \) is valid. The base must be positive and cannot be equal to 1:

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

The condition \( x > -4 \) is less strict than \( x > \frac{1}{2} \), so we will use \( x > \frac{1}{2} \).

### Step 3: Determine when the logarithm is non-positive

Now we need to solve:

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

This inequality holds when:

1. \( \log_2 \left( \frac{2x-1}{3+x} \right) \leq 1 \) when \( \frac{x+4}{2} > 1 \) (which is true for \( x > -2 \)).
2. \( \log_2 \left( \frac{2x-1}{3+x} \right) \geq 1 \) when \( \frac{x+4}{2} < 1 \) (not applicable here since \( x > \frac{1}{2} \)).

So we solve:

\[ \frac{2x-1}{3+x} \leq 2 \]

Cross-multiplying gives:

\[ 2x - 1 \leq 2(3 + x) \]

This simplifies to:

\[ 2x - 1 \leq 6 + 2x \]

Subtracting \( 2x \) from both sides:

\[ -1 \leq 6 \]

This is always true, so we need to check the other condition \( \frac{2x-1}{3+x} > 0 \) which we already established.

### Conclusion

The only condition that restricts \( x \) is \( x > \frac{1}{2} \). Therefore, the domain of the function is:

\[ \boxed{( \frac{1}{2}, \infty )} \]