If n = (2017)! then what is \(\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + \ldots + \frac{1}{{{{\log }_{2017}}n}}\) equal to?
1
Understanding the Logarithm Sum Problem
The question asks us to evaluate a sum of terms involving logarithms. We are given that \(n = (2017)!\) and we need to find the value of the expression:
\( \frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + \ldots + \frac{1}{{{{\log }_{2017}}n}} \)
Let's break down this problem using fundamental logarithm properties.
Applying Logarithm Properties
A crucial property of logarithms is the change of base formula or its inverse form, which states that \(\frac{1}{{{{\log }_b}a}} = {\log _a}b\). We can apply this property to each term in the given sum.
- The first term is \(\frac{1}{{{{\log }_2}n}}\). Using the property, this becomes \({\log _n}2\).
- The second term is \(\frac{1}{{{{\log }_3}n}}\). Using the property, this becomes \({\log _n}3\).
- The third term is \(\frac{1}{{{{\log }_4}n}}\). Using the property, this becomes \({\log _n}4\).
- This pattern continues up to the last term, which is \(\frac{1}{{{{\log }_{2017}}n}}\). Using the property, this becomes \({\log _n}2017\).
So, the original sum can be rewritten as:
\( {\log _n}2 + {\log _n}3 + {\log _n}4 + \ldots + {\log _n}2017 \)
Simplifying the Sum of Logarithms
Now we have a sum of logarithms with the same base, which is \(n\). Another key property of logarithms states that the sum of logarithms with the same base is the logarithm of the product of their arguments: \({\log _b}x + {\log _b}y = {\log _b}(xy)\). We can extend this property to the entire sum:
\( {\log _n}2 + {\log _n}3 + {\log _n}4 + \ldots + {\log _n}2017 = {\log _n}(2 \times 3 \times 4 \times \ldots \times 2017) \)
Evaluating the Product
The product inside the logarithm is \(2 \times 3 \times 4 \times \ldots \times 2017\). This is almost the definition of a factorial. The factorial of a positive integer \(k\), denoted by \(k!\), is the product of all positive integers up to \(k\): \(k! = 1 \times 2 \times 3 \times \ldots \times k\). The product \(2 \times 3 \times 4 \times \ldots \times 2017\) is the product of integers from 2 up to 2017. To make it a full factorial \( (2017)! \), we just need to include the factor of 1. Since multiplying by 1 does not change the value, we can write:
\( 2 \times 3 \times 4 \times \ldots \times 2017 = 1 \times 2 \times 3 \times 4 \times \ldots \times 2017 = (2017)! \)
So, the expression simplifies to:
\( {\log _n}((2017)!) \)
Final Calculation
We are given that \(n = (2017)!\). Let's substitute this value of \(n\) into the expression:
\( {\log _{(2017)!}}((2017)!) \)
Finally, we use the property that \({\log _b}b = 1\) for any base \(b > 0\) and \(b \ne 1\). In our case, the base is \((2017)!\) and the argument is also \((2017)!\). Since \((2017)!\) is a very large positive integer (greater than 1), this property applies.
Therefore,
\( {\log _{(2017)!}}((2017)!) = 1 \)
The value of the given expression is 1.
Summary of Steps
Here is a summary of the steps taken to solve the problem:
- Recognize the given expression as a sum involving logarithms with changing bases but a constant argument \(n = (2017)!\).
- Apply the logarithm property \(\frac{1}{{{{\log }_b}a}} = {\log _a}b\) to rewrite each term with base \(n\).
- Rewrite the sum using the property \({\log _n}b_1 + {\log _n}b_2 + \ldots = {\log _n}(b_1 \times b_2 \times \ldots)\).
- Identify the product \(2 \times 3 \times \ldots \times 2017\) as \((2017)!\).
- Substitute \(n = (2017)!\) into the simplified logarithmic expression.
- Use the property \({\log _b}b = 1\) to find the final value.
| Expression Step | Mathematical Form | Logarithm Property Used |
|---|---|---|
| Original Sum | \( \sum_{k=2}^{2017} \frac{1}{{{{\log }_k}n}} \) | Given |
| Apply Inverse Property | \( \sum_{k=2}^{2017} {{\log }_n}k \) | \( \frac{1}{{{{\log }_b}a}} = {\log _a}b \) |
| Apply Sum Property | \( {\log _n}(2 \times 3 \times \ldots \times 2017) \) | \( {\log _b}x + {\log _b}y = {\log _b}(xy) \) |
| Evaluate Product | \( {\log _n}((2017)!) \) | Definition of Factorial |
| Substitute \(n=(2017)!\) | \( {\log _{(2017)!}}((2017)!) \) | Given \(n\) |
| Final Evaluation | \( 1 \) | \( {\log _b}b = 1 \) |
Conclusion on the Logarithm Problem
By applying standard logarithm properties and understanding the definition of a factorial, the complex-looking sum simplifies neatly to a logarithm where the base and argument are identical, resulting in a value of 1.
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Logarithm Base Change Inverse | \( \frac{1}{{{{\log }_b}a}} = {\log _a}b \) | Used to change base from \(k\) to \(n\) for each term. |
| Sum of Logarithms | \( {\log _b}x + {\log _b}y = {\log _b}(xy) \) | Used to combine the sum into a single logarithm of a product. |
| Factorial | \( k! = 1 \times 2 \times \ldots \times k \) | Used to recognize the product \(2 \times 3 \times \ldots \times 2017\) as \((2017)!\). |
| Logarithm of Base Itself | \( {\log _b}b = 1 \) | Used for the final step to evaluate \({\log _{(2017)!}}((2017)!)\). |
Additional Information: Logarithms and Factorials
Logarithms are the inverse operations to exponentiation. They are extremely useful for simplifying calculations involving multiplication, division, and exponents, especially with large numbers. The expression we evaluated demonstrates how logarithm properties can transform a sum into a product inside the logarithm.
Factorials, denoted by \(n!\), grow very rapidly. \( (2017)! \) is an incredibly large number. Problems involving factorials often test your understanding of their definition and how they might relate to other mathematical concepts like logarithms, as seen in this problem.
Understanding these core mathematical concepts and their properties is essential for solving such problems in competitive exams and advanced mathematics.
Similar Questions
Let y = [x + 1], -4 < x < -3 where [.] is the greatest integer function. What is the derivative of y with respect to x at x = -3.5?
The value of x, satisfying the equation \(log_{cos x} ~sin x = 1\) , where \(0<x<\dfrac{\pi}{2}\) , is
If \({{x}^{{{\log }_{7}}x}}>7\) where x > 0, then which one of the following is correct?
If f(x) = log 10 (1 + x), then what is 4f(4) + 5f(1) – log 10 2 equal to?
A function f defined by f(x) = In \(\left( {\sqrt {{x^2} + 1} - x} \right)\) is
If f(x) = 3 1+x , then f(x) f(y) f(z) is equal to
If (0.2) x= 2 and log 10 2 = 0.3010, the what is the value of x to the nearest tenth?
IF x + log 15 (1 + 3 x) = x log 15 5 + log 15 12, where x is an integer, then what is x equal to?
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)?
What is the value of log 927 + log 832?
Important Questions from Special Functions
- The function $f(x) = [2x]$ where $[x]$ is the greatest integer function, is continuous at
The number of real solutions of equation x 2 - 3 |x| + 2 = 0 is:
If ϕ is the Euler’s Totient function, then ϕ(92) is:
Consider the linear congruence 6 x ≡ 3 (mod 9). Then the incongruent solutions modulo 9 of this congruence are:
If log10(x2 - 6x + 45) = 2, then the value of x are:
