What is \(\frac{1}{{{{\log }_2}N}} + \frac{1}{{{{\log }_3}N}} + \frac{1}{{{{\log }_4}N}} + \ldots + \frac{1}{{{{\log }_{100}}N\;}}\;\) equal to (N ≠ 1)?
Simplifying the Sum of Reciprocal Logarithms
The question asks us to simplify a given sum involving reciprocals of logarithms with different bases but the same argument N (where N ≠ 1).
The expression is:
\(\frac{1}{{{{\log }_2}N}} + \frac{1}{{{{\log }_3}N}} + \frac{1}{{{{\log }_4}N}} + \ldots + \frac{1}{{{{\log }_{100}}N\;}}\)
To simplify this sum, we need to use a fundamental property of logarithms known as the change of base property or reciprocal identity. The identity states that:
Logarithm Reciprocal Identity:
\(\frac{1}{{{\log _b}a}} = {\log _a}b\)
This identity allows us to swap the base and the argument of a logarithm when taking its reciprocal. Let's apply this property to each term in the given sum. Here, 'b' corresponds to the base (2, 3, 4, ..., 100) and 'a' corresponds to the argument N.
- The first term: \(\frac{1}{{{{\log }_2}N}} = {\log _N}2\)
- The second term: \(\frac{1}{{{{\log }_3}N}} = {\log _N}3\)
- The third term: \(\frac{1}{{{{\log }_4}N}} = {\log _N}4\)
- ...
- The last term: \(\frac{1}{{{{\log }_{100}}N}} = {\log _N}100\)
Now, substitute these back into the original sum:
The sum becomes: \({\log _N}2 + {\log _N}3 + {\log _N}4 + \ldots + {\log _N}100\)
This is a sum of logarithms with the same base N. We can use another important property of logarithms: the product rule.
Logarithm Product Rule:
\({\log _b}x + {\log _b}y = {\log _b}(xy)\)
This rule can be extended to a sum of multiple logarithms with the same base. The sum of logarithms is equal to the logarithm of the product of their arguments.
Applying the product rule to our sum:
\({\log _N}2 + {\log _N}3 + {\log _N}4 + \ldots + {\log _N}100 = {\log _N}(2 \times 3 \times 4 \times \ldots \times 100)\)
The product inside the logarithm is \(2 \times 3 \times 4 \times \ldots \times 100\). This product is the factorial of 100, excluding the multiplication by 1 (which doesn't change the product). The product \(1 \times 2 \times 3 \times \ldots \times 100\) is denoted as \(100!\).
So, \(2 \times 3 \times 4 \times \ldots \times 100 = 100!\)
Therefore, the sum simplifies to:
\({\log _N}(100!)\)
Now, let's look at the given options. The options are in the form \(\frac{1}{{{\log _k}N}}\). We need to express our result \({\log _N}(100!)\) in this form.
We can use the reciprocal identity \(\frac{1}{{{\log _b}a}} = {\log _a}b\) again, but this time in the reverse direction. Here, 'a' is N and 'b' is \(100!\). So, we have:
\({\log _N}(100!) = \frac{1}{{{\log }_{100!}}N}\)
Comparing this result with the given options:
| Option | Expression | Match? |
|---|---|---|
| 1 | \(\frac{1}{{{{\log }_{100!}}N}}\) | Yes |
| 2 | \(\frac{1}{{{{\log }_{99!}}N}}\) | No |
| 3 | \(\frac{{99}}{{{{\log }_{100!}}N}}\) | No |
| 4 | \(\frac{{99}}{{{{\log }_{99!}}N}}\) | No |
The simplified expression is \(\frac{1}{{{{\log }_{100!}}N}}\), which matches Option 1.
Revision Table: Key Logarithm Properties Used
| Property Name | Formula | Application in Solution |
|---|---|---|
| Reciprocal Identity (Change of Base) | \(\frac{1}{{{\log _b}a}} = {\log _a}b\) | Used to convert each term \(\frac{1}{{{\log _b}N}}\) to \({\log _N}b\). Also used to convert the final result \({\log _N}(100!)\) back to the required form. |
| Product Rule | \({\log _b}x + {\log _b}y = {\log _b}(xy)\) | Used to combine the sum \({\log _N}2 + \ldots + {\log _N}100\) into a single logarithm \({\log _N}(2 \times \ldots \times 100)\). |
Additional Information on Logarithms and Factorials
What is a Logarithm?
A logarithm is the inverse operation to exponentiation. This means the logarithm of a number N with respect to a base b is the exponent by which b must be raised to produce N. It's written as \({\log _b}N = x\), which is equivalent to \(b^x = N\).
Why is N ≠ 1 important?
In the definition of a logarithm \({\log _b}N\), the base b must be a positive number other than 1, and the argument N must be a positive number. The condition N ≠ 1 is given in the question because if N were 1, \({\log _b}1 = 0\) for any valid base b. This would make the denominators in the original sum zero, leading to undefined terms. Thus, N ≠ 1 ensures the logarithms are well-defined and non-zero.
What is a Factorial?
The factorial of a non-negative integer n, denoted by \(n!\), is the product of all positive integers less than or equal to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). The expression \(2 \times 3 \times \ldots \times 100\) is indeed \(100!\) because multiplying by 1 does not change the product.
Understanding these core concepts and properties is crucial for solving problems involving logarithmic expressions.
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