`a=log_5 3+log_7 5+log_9 7`

Options

A. `ain[3//2,oo)`
B. `ain[(1)/(2^(1//3),oo)`
C. `ain[(3)/(2^(1//3),oo)`
D. none of these.

Explanation

To solve the problem \( a = \log_5 3 + \log_7 5 + \log_9 7 \) and find the interval of \( a \), we can use the Arithmetic Mean - Geometric Mean (AM-GM) inequality. Here’s a step-by-step solution:

### Step 1: Apply the AM-GM Inequality
According to the AM-GM inequality for three positive numbers \( x_1, x_2, x_3 \):
\[ \frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3} \]
In our case, let:
\[ x_1 = \log_5 3, \quad x_2 = \log_7 5, \quad x_3 = \log_9 7 \]
Thus, we have:
\[ \frac{\log_5 3 + \log_7 5 + \log_9 7}{3} \geq \sqrt[3]{\log_5 3 \cdot \log_7 5 \cdot \log_9 7} \]
This implies:
\[ \frac{a}{3} \geq \sqrt[3]{\log_5 3 \cdot \log_7 5 \cdot \log_9 7} \]

### Step 2: Rewrite the Logarithms
Using the change of base formula, we can rewrite the logarithms:
\[ \log_5 3 = \frac{\log 3}{\log 5}, \quad \log_7 5 = \frac{\log 5}{\log 7}, \quad \log_9 7 = \frac{\log 7}{\log 9} \]
Thus, we can express \( a \) as:
\[ a \geq 3 \sqrt[3]{\frac{\log 3}{\log 5} \cdot \frac{\log 5}{\log 7} \cdot \frac{\log 7}{\log 9}} \]

### Step 3: Simplify the Expression
Notice that the \( \log 5 \) and \( \log 7 \) terms will cancel out:
\[ \log 5 \text{ cancels with } \log 5 \text{ and } \log 7 \text{ cancels with } \log 7 \]
Thus, we have:
\[ a \geq 3 \sqrt[3]{\frac{\log 3}{\log 9}} = 3 \sqrt[3]{\frac{\log 3}{\log 3^2}} = 3 \sqrt[3]{\frac{\log 3}{2 \log 3}} = 3 \sqrt[3]{\frac{1}{2}} = \frac{3}{2^{1/3}} \]

### Step 4: Conclusion
Therefore, we conclude that:
\[ a \geq \frac{3}{2^{1/3}} \]
This means that the interval of \( a \) is:
\[ a \in \left[\frac{3}{2^{1/3}}, \infty\right) \]

### Final Answer
The interval of \( a \) is:
\[ \boxed{\left[\frac{3}{2^{1/3}}, \infty\right)} \]