Write the domain of the following real functions
`log(sqrt(x-4)+sqrt(6-x))`

Answer

To find the domain of the function \( f(x) = \log(\sqrt{x-4} + \sqrt{6-x}) \), we need to ensure that the expression inside the logarithm is positive, as logarithms are only defined for positive arguments.

### Step-by-Step Solution:

1. **Set up the inequality for the logarithm**:
The logarithmic function is defined when its argument is greater than zero. Therefore, we need:
\[ \sqrt{x-4} + \sqrt{6-x} > 0 \]

2. **Analyze the square root terms**:
Each square root must be defined, which means:
- For \( \sqrt{x-4} \) to be defined, we need:
\[ x - 4 \geq 0 \implies x \geq 4 \]
- For \( \sqrt{6-x} \) to be defined, we need:
\[ 6 - x \geq 0 \implies x \leq 6 \]

3. **Combine the inequalities**:
From the above conditions, we have:
\[ 4 \leq x \leq 6 \]
This means \( x \) must be in the interval \([4, 6]\).

4. **Check the endpoints**:
- At \( x = 4 \):
\[ \sqrt{4-4} + \sqrt{6-4} = 0 + \sqrt{2} > 0 \]
- At \( x = 6 \):
\[ \sqrt{6-4} + \sqrt{6-6} = \sqrt{2} + 0 > 0 \]
Both endpoints satisfy the condition that the argument of the logarithm is positive.

5. **Conclusion**:
Therefore, the domain of the function \( f(x) = \log(\sqrt{x-4} + \sqrt{6-x}) \) is:
\[ \text{Domain} = [4, 6] \]