If `x=prod_(n=1)^(2000)n,` then the value of the expression, `(1)/((1)/(log_(2)x)+(1)/(log_(3)x)+.........+(1)/(log_(2000)x))is..........`

Answer

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

\[ \frac{1}{\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \ldots + \frac{1}{\log_{2000} x}} \]

where \( x = \prod_{n=1}^{2000} n \).

### Step 1: Understanding \( x \)
We know that \( x \) is the product of the first 2000 natural numbers, which is equal to \( 2000! \).

### Step 2: Expressing the logarithm
Using the change of base formula for logarithms, we can express \( \log_b a \) as \( \frac{\log a}{\log b} \). Thus, we can rewrite each term in the denominator:

\[ \log_b x = \log_b (2000!) = \frac{\log (2000!)}{\log b} \]

### Step 3: Rewriting the denominator
Now substituting this into our expression, we have:

\[ \frac{1}{\frac{1}{\log_2 (2000!)} + \frac{1}{\log_3 (2000!)} + \ldots + \frac{1}{\log_{2000} (2000!)}} \]

This becomes:

\[ \frac{1}{\frac{\log 2}{\log (2000!)} + \frac{\log 3}{\log (2000!)} + \ldots + \frac{\log 2000}{\log (2000!)}} \]

### Step 4: Factoring out \( \log (2000!) \)
We can factor out \( \frac{1}{\log (2000!)} \):

\[ = \frac{1}{\frac{1}{\log (2000!)} \left( \log 2 + \log 3 + \ldots + \log 2000 \right)} \]

### Step 5: Simplifying the expression
This simplifies to:

\[ \frac{\log (2000!)}{\log 2 + \log 3 + \ldots + \log 2000} \]

Using the property of logarithms that states \( \log a + \log b = \log (a \cdot b) \), we can rewrite the denominator:

\[ \log 2 + \log 3 + \ldots + \log 2000 = \log (2 \cdot 3 \cdot \ldots \cdot 2000) = \log (2000!) \]

### Step 6: Final simplification
Now substituting this back into our expression gives:

\[ \frac{\log (2000!)}{\log (2000!)} = 1 \]

### Conclusion
Thus, the value of the expression is:

\[ \boxed{1} \]