Consider the system of equations
` log_(3)(log_(2)x)+log_(1//3)(log_(1//2)y) =1 and xy^(2) = 9`.
The value of 1/y lies in the interval

Options

A. `(5, 7)`
B. `(7, 10)`
C. `(11, 15)`
D. `(25, 30)`

Explanation

To solve the given system of equations:

1. **Equations**:
\[ \log_{3}(\log_{2} x) + \log_{\frac{1}{3}}(\log_{\frac{1}{2}} y) = 1 \]
\[ xy^{2} = 9 \]

2. **Rewriting the first equation**:
We know that \(\log_{\frac{1}{3}} a = -\log_{3} a\). Therefore, we can rewrite the first equation as:
\[ \log_{3}(\log_{2} x) - \log_{3}(\log_{\frac{1}{2}} y) = 1 \]

3. **Using the property of logarithms**:
The property \(\log_{a} b - \log_{a} c = \log_{a} \left(\frac{b}{c}\right)\) allows us to combine the logarithms:
\[ \log_{3} \left(\frac{\log_{2} x}{\log_{\frac{1}{2}} y}\right) = 1 \]

4. **Exponentiating both sides**:
This implies:
\[ \frac{\log_{2} x}{\log_{\frac{1}{2}} y} = 3 \]

5. **Substituting \(\log_{\frac{1}{2}} y\)**:
We know that \(\log_{\frac{1}{2}} y = -\log_{2} y\). Thus, we can rewrite the equation as:
\[ \frac{\log_{2} x}{-\log_{2} y} = 3 \]
This simplifies to:
\[ \log_{2} x = -3 \log_{2} y \]

6. **Expressing \(x\) in terms of \(y\)**:
From the above equation, we can express \(x\) as:
\[ x = y^{-3} \]

7. **Substituting into the second equation**:
Now, substituting \(x = y^{-3}\) into the second equation \(xy^{2} = 9\):
\[ (y^{-3})y^{2} = 9 \]
This simplifies to:
\[ \frac{y^{2}}{y^{3}} = 9 \quad \Rightarrow \quad \frac{1}{y} = 9 \]

8. **Finding \(y\)**:
Therefore, we have:
\[ y = \frac{1}{9} \]

9. **Finding \(\frac{1}{y}\)**:
Now, we need to find \(\frac{1}{y}\):
\[ \frac{1}{y} = 9 \]

10. **Determining the interval**:
We need to check in which interval \(9\) lies:
- \(5\) to \(7\) (No)
- \(7\) to \(10\) (Yes)
- \(11\) to \(15\) (No)
- \(25\) to \(30\) (No)

Thus, the value of \(\frac{1}{y}\) lies in the interval \(7\) to \(10\).