Understanding the Logarithm Sum Problem

We are asked to find the value of a sum involving logarithms with different bases but a common argument, \(n\). The expression is:

\(\rm \dfrac{1}{log_2n}+\dfrac{1}{log_3n}+\dfrac{1}{log_4n}+{.....}+\dfrac{1}{log_{100}n}\)

We are also given that \(n = 100!\). To solve this logarithm problem, we need to use fundamental properties of logarithms.

Applying Logarithm Properties for Simplification

The expression involves terms of the form \(\frac{1}{\log_b a}\). There is a useful property of logarithms that relates this form to a logarithm with the base and argument swapped:

Logarithm Reciprocal Property: \(\dfrac{1}{\log_b a} = \log_a b\)

Using this property, we can rewrite each term in the given sum. For example:

• \(\dfrac{1}{\log_2 n} = \log_n 2\)
• \(\dfrac{1}{\log_3 n} = \log_n 3\)
• \(\dfrac{1}{\log_4 n} = \log_n 4\)
• ...
• \(\dfrac{1}{\log_{100} n} = \log_n 100\)

So, the sum can be rewritten as:

\(\log_n 2 + \log_n 3 + \log_n 4 + {.....} + \log_n 100\)

Now we have a sum of logarithms with the same base, \(n\). There is another key logarithm property for the sum of logarithms:

Logarithm Sum Property: \(\log_b x + \log_b y = \log_b (xy)\)

We can extend this property to a sum of multiple terms:

\(\log_b x_1 + \log_b x_2 + {.....} + \log_b x_k = \log_b (x_1 \times x_2 \times {.....} \times x_k)\)

Applying this property to our sum:

\(\log_n 2 + \log_n 3 + \log_n 4 + {.....} + \log_n 100 = \log_n (2 \times 3 \times 4 \times {.....} \times 100)\)

The product inside the logarithm is \(2 \times 3 \times 4 \times {.....} \times 100\). This is the product of all integers from 2 up to 100. This product is equal to \(100!\), because \(100! = 1 \times 2 \times 3 \times {.....} \times 100\), and multiplying by 1 does not change the product.

So, the expression simplifies to \(\log_n (100!)\).

Calculating the Final Value

We are given that \(n = 100!\). We can substitute this value of \(n\) into the simplified expression \(\log_n (100!)\):

\(\log_{100!} (100!)\)

Finally, we use the basic definition of a logarithm or the property \(\log_b b = 1\).

Logarithm Identity Property: \(\log_b b = 1\), for \(b > 0\) and \(b \neq 1\).

Since the base is \(100!\) and the argument is also \(100!\), and \(100!\) is a large positive number not equal to 1, we have:

\(\log_{100!} (100!) = 1\)

Thus, the value of the given logarithm sum is 1.

Revision Table: Key Logarithm Concepts

Additional Information on Factorials and Logarithms

Factorial (n!): For a positive integer \(n\), the factorial \(n!\) is the product of all positive integers less than or equal to \(n\). \(n! = n \times (n-1) \times {.....} \times 2 \times 1\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). \(0!\) is defined as 1.

Logarithm: A logarithm is the inverse operation to exponentiation. The logarithm of a number \(x\) with respect to base \(b\) is the exponent to which \(b\) must be raised to produce \(x\). It is written as \(\log_b x\). For example, \(\log_{10} 100 = 2\) because \(10^2 = 100\).

Understanding these basic definitions and properties is crucial for solving problems involving logarithmic expressions like the one presented.