If `x_n > x_(n-1) > ..........> x_3 > x_1 > 1.` then the value of `log_(x_1) [log_(x _2) {log_(x_3).........log_(x_n) (x_n)^(x_(r=i))}]`

Options

A. 0
B. 1
C. 2
D. undefined

Answer

To solve the problem, we need to evaluate the expression:

\[ \log_{x_1} \left[ \log_{x_2} \left\{ \log_{x_3} \ldots \log_{x_n} (x_n)^{x_{r=i}} \right\} \right] \]

Given that \( x_n > x_{n-1} > \ldots > x_3 > x_1 > 1 \).

### Step-by-step Solution:

1. **Start with the innermost logarithm**:
We begin with the innermost part of the expression, which is \( \log_{x_n} (x_n)^{x_{r=i}} \).
By using the property of logarithms, \( \log_b (a^c) = c \cdot \log_b a \), we can simplify this:
\[ \log_{x_n} (x_n)^{x_{r=i}} = x_{r=i} \cdot \log_{x_n} (x_n) = x_{r=i} \cdot 1 = x_{r=i} \]

2. **Substituting back**:
Now we substitute \( x_{r=i} \) back into the next logarithm:
\[ \log_{x_{n-1}}(x_{r=i}) \]
Since \( x_{r=i} \) is less than \( x_{n-1} \) (because \( x_n > x_{n-1} > x_1 > 1 \)), this logarithm will yield a positive value.

3. **Continuing the process**:
Continuing this process, we substitute the result back into the next logarithm:
\[ \log_{x_{n-2}}(\log_{x_{n-1}}(x_{r=i})) \]
Again, since \( x_{n-2} > x_{n-1} > x_{r=i} \), this will also yield a positive value.

4. **Repeating until the outermost logarithm**:
We repeat this process until we reach the outermost logarithm:
\[ \log_{x_1}(\log_{x_2}(\log_{x_3}(\ldots))) \]
Each step will yield a positive value, and since all bases \( x_1, x_2, \ldots, x_n \) are greater than 1, the logarithm of any positive number will also be defined.

5. **Final evaluation**:
Ultimately, if we continue this process, we will find that:
\[ \log_{x_1}(\text{some positive value}) \rightarrow 1 \]
This is because as we keep applying logarithms, we eventually reach the point where the logarithm of the base equals the base, yielding a value of 1.

Thus, the final value of the entire expression is:

\[ \boxed{1} \]