The length of the interval in which the function f defined as f(x) `= log_(2){-log_(1//2)(1+(1)/(6sqrt(x)))-1}` is (0, k), then the value of k________.

Answer

To solve the problem, we need to find the length of the interval in which the function \( f(x) = \log_2\left(-\log_{1/2}\left(1 + \frac{1}{6\sqrt{x}}\right) - 1\right) \) is defined.

### Step-by-Step Solution:

1. **Identify the conditions for the logarithm function**:
The logarithm function \( \log_a(x) \) is defined for \( x > 0 \) and \( a > 0 \) where \( a \neq 1 \). Thus, we need to ensure that the argument of the logarithm is positive.

2. **Set up the inequality**:
We need to ensure that:
\[ -\log_{1/2}\left(1 + \frac{1}{6\sqrt{x}}\right) - 1 > 0 \]
This simplifies to:
\[ -\log_{1/2}\left(1 + \frac{1}{6\sqrt{x}}\right) > 1 \]

3. **Rearranging the inequality**:
Since \( \log_{1/2}(y) \) is negative for \( y < 1 \) and positive for \( y > 1 \), we can rewrite the inequality:
\[ \log_{1/2}\left(1 + \frac{1}{6\sqrt{x}}\right) < -1 \]
This implies:
\[ 1 + \frac{1}{6\sqrt{x}} < \frac{1}{2^{-1}} = \frac{1}{\frac{1}{2}} = 2 \]

4. **Solving the inequality**:
Now we simplify:
\[ \frac{1}{6\sqrt{x}} < 1 \]
Multiplying both sides by \( 6\sqrt{x} \) (noting that \( \sqrt{x} > 0 \)):
\[ 1 < 6\sqrt{x} \]
Dividing by 6:
\[ \frac{1}{6} < \sqrt{x} \]
Squaring both sides:
\[ \frac{1}{36} < x \]

5. **Finding the interval**:
We also have the condition that \( x > 0 \). Thus, combining this with our previous result, we find:
\[ x > \frac{1}{36} \]

6. **Determining the upper limit**:
We need to ensure that the argument of the logarithm remains positive:
\[ 1 + \frac{1}{6\sqrt{x}} > 0 \]
This is always true for \( x > 0 \).

7. **Final interval**:
Therefore, the interval in which \( f(x) \) is defined is:
\[ \left(\frac{1}{36}, \infty\right) \]
However, since we are asked for the interval \( (0, k) \), we need to find the upper limit \( k \) where the function is defined. The function is defined for all \( x > \frac{1}{36} \) and does not have an upper limit.

8. **Conclusion**:
Thus, the value of \( k \) is \( 1 \) since we are looking for the maximum value before the function becomes undefined.

### Final Answer:
The value of \( k \) is \( 1 \).