If `x_(n)gt1` for all `n in N`, then the minimum value of the expression
`log_(x_(2))x_(1)+log_(x_(3))x_(2)+...+log_(x_(n))x_(n-1)+log_(x_(1))x_(n)` is

Options

A. 0
B. 1
C. 2
D. none of these

Explanation

To find the minimum value of the expression

\[ \log_{x_2} x_1 + \log_{x_3} x_2 + \ldots + \log_{x_n} x_{n-1} + \log_{x_1} x_n \]

given that \( x_n > 1 \) for all \( n \in \mathbb{N} \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Identify the Terms**:
The expression consists of \( n \) logarithmic terms:
\[ \log_{x_2} x_1, \log_{x_3} x_2, \ldots, \log_{x_n} x_{n-1}, \log_{x_1} x_n \]

2. **Apply AM-GM Inequality**:
According to the AM-GM inequality:
\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n} \]
where \( a_i \) are the terms we are considering. Here, we can set:
\[ a_1 = \log_{x_2} x_1, \quad a_2 = \log_{x_3} x_2, \quad \ldots, \quad a_n = \log_{x_1} x_n \]

3. **Calculate the Arithmetic Mean**:
The arithmetic mean of these terms is:
\[ \frac{\log_{x_2} x_1 + \log_{x_3} x_2 + \ldots + \log_{x_1} x_n}{n} \]

4. **Express Logarithms in Terms of Natural Logarithms**:
Using the change of base formula, we can express each logarithm:
\[ \log_{x_k} x_{k-1} = \frac{\log x_{k-1}}{\log x_k} \]
Therefore, the product of all terms becomes:
\[ \log_{x_2} x_1 \cdot \log_{x_3} x_2 \cdots \log_{x_1} x_n = \frac{\log x_1}{\log x_2} \cdot \frac{\log x_2}{\log x_3} \cdots \frac{\log x_n}{\log x_1} \]

5. **Simplify the Product**:
Notice that in the product, all logarithmic terms will cancel out:
\[ = \frac{\log x_1}{\log x_1} = 1 \]

6. **Apply AM-GM Result**:
Thus, we have:
\[ \frac{\log_{x_2} x_1 + \log_{x_3} x_2 + \ldots + \log_{x_1} x_n}{n} \geq \sqrt[n]{1} = 1 \]

7. **Multiply by n**:
Therefore, multiplying both sides by \( n \):
\[ \log_{x_2} x_1 + \log_{x_3} x_2 + \ldots + \log_{x_1} x_n \geq n \]

8. **Conclusion**:
The minimum value of the expression is \( n \).

### Final Answer:
The minimum value of the expression is \( n \).