In n = 10!, then what is the value of the following?
`(1)/(log_(2)n) +(1)/(log_(3)n) +(1)/(log_(4)n)+…..+ (1)/(log_(10)n)`

Options

A. 0
B. 1
C. 2
D. 3

Answer

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

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \ldots + \frac{1}{\log_{10} n} \]

where \( n = 10! \).

### Step 1: Rewrite the logarithmic terms

Using the property of logarithms that states:

\[ \log_b a = \frac{1}{\log_a b} \]

we can rewrite each term in the sum:

\[ \frac{1}{\log_k n} = \log_n k \]

Thus, we can rewrite the entire expression as:

\[ \log_n 2 + \log_n 3 + \log_n 4 + \ldots + \log_n 10 \]

### Step 2: Combine the logarithmic terms

Using the property of logarithms that states:

\[ \log_a b + \log_a c = \log_a (bc) \]

we can combine the terms:

\[ \log_n (2 \times 3 \times 4 \times \ldots \times 10) \]

### Step 3: Calculate the product

The product \( 2 \times 3 \times 4 \times \ldots \times 10 \) is equal to \( 10! \) (since it is the product of all integers from 1 to 10, excluding 1):

\[ 2 \times 3 \times 4 \times \ldots \times 10 = \frac{10!}{1} = 10! \]

### Step 4: Substitute back into the logarithm

Now we can substitute this back into our logarithmic expression:

\[ \log_n (10!) \]

### Step 5: Substitute \( n = 10! \)

Since \( n = 10! \), we have:

\[ \log_{10!} (10!) \]

### Step 6: Evaluate the logarithm

Using the property that \( \log_a a = 1 \):

\[ \log_{10!} (10!) = 1 \]

### Final Answer

Thus, the value of the original expression is:

\[ \boxed{1} \]