The domain of `y=log_(10)(sqrt(6-x)+sqrt(x-4))` is :

Options

A. R
B. `R-{0}`
C. `[6, oo)`
D. `(4, 6)`

Answer

To determine the domain of the function \( y = \log_{10}(\sqrt{6-x} + \sqrt{x-4}) \), we need to ensure that the expression inside the logarithm is positive, as the logarithm is only defined for positive values. Additionally, we must ensure that the square roots are defined.

### Step-by-Step Solution:

1. **Identify conditions for the square roots:**
- The term \( \sqrt{6-x} \) is defined when \( 6-x \geq 0 \).
- The term \( \sqrt{x-4} \) is defined when \( x-4 \geq 0 \).

2. **Set up inequalities:**
- From \( 6-x \geq 0 \):
\[ 6 \geq x \quad \Rightarrow \quad x \leq 6 \]
- From \( x-4 \geq 0 \):
\[ x \geq 4 \]

3. **Combine the inequalities:**
- We have two inequalities:
\[ 4 \leq x \leq 6 \]
- This means \( x \) must be in the interval \( [4, 6] \).

4. **Check the expression inside the logarithm:**
- We need \( \sqrt{6-x} + \sqrt{x-4} > 0 \).
- Since both square roots are non-negative, the sum is positive as long as both square roots are defined and at least one of them is positive.
- At \( x = 4 \):
\[ \sqrt{6-4} + \sqrt{4-4} = \sqrt{2} + 0 > 0 \]
- At \( x = 6 \):
\[ \sqrt{6-6} + \sqrt{6-4} = 0 + \sqrt{2} > 0 \]
- For any \( x \) in the interval \( (4, 6) \), both square roots will be positive.

5. **Conclusion:**
- Therefore, the domain of the function \( y = \log_{10}(\sqrt{6-x} + \sqrt{x-4}) \) is:
\[ [4, 6] \]

### Final Answer:
The domain of \( y = \log_{10}(\sqrt{6-x} + \sqrt{x-4}) \) is \( [4, 6] \).