If `n=1999!` then `sum_(x=1)^(1999) log_n x=`

Options

A. 1
B. 0
C. `root(1999)(1999)`
D. `-1`

Explanation

To solve the problem \( \sum_{x=1}^{1999} \log_n x \) where \( n = 1999! \), we can follow these steps:

### Step 1: Rewrite the Summation
We start with the summation:
\[ \sum_{x=1}^{1999} \log_n x = \log_n 1 + \log_n 2 + \log_n 3 + \ldots + \log_n 1999 \]

### Step 2: Use the Property of Logarithms
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can combine the terms:
\[ \sum_{x=1}^{1999} \log_n x = \log_n (1 \cdot 2 \cdot 3 \cdots \cdot 1999) = \log_n (1999!) \]

### Step 3: Substitute \( n \)
Since we know that \( n = 1999! \), we can substitute \( n \) into the logarithm:
\[ \log_n (1999!) = \log_{1999!} (1999!) \]

### Step 4: Apply the Logarithm Identity
Using the identity \( \log_a a = 1 \), we find:
\[ \log_{1999!} (1999!) = 1 \]

### Conclusion
Thus, the value of the summation \( \sum_{x=1}^{1999} \log_n x \) is:
\[ \boxed{1} \]