Problem 47 - Calculus Single Variable

Question

Find \(p\) such that \(e^{p}=e \cdot e^{2} \cdot e^{3} \ldots e^{100}\). Find \(q\) so that \(\ln (q)=\sum_{n=1}^{100} \ln (n)\)

Step-by-Step Solution

Answer

\( p = 5050 \), \( q = 100! \).

Step 1: Simplify the Exponential Equation

The given problem is to find \( p \) such that \( e^p = e \cdot e^2 \cdot e^3 \ldots e^{100} \). Notice that the right side is a product of exponentials. We can rewrite this as a single exponential by adding the exponents: \( e^p = e^{1+2+3+\ldots+100} \).

Step 2: Calculate the Sum of Integers

The expression inside the exponent is the sum of integers from 1 to 100. This is an arithmetic series, which can be computed using the formula for the sum of the first \( n \) integers: \( \frac{n(n+1)}{2} \). So we find the sum as \( \frac{100 \cdot (100+1)}{2} \).

Step 3: Compute the Sum

Calculate the value of the sum: \( \frac{100 \cdot 101}{2} = 5050 \). Therefore, \( e^p = e^{5050} \).

Step 4: Solve for p

Since the base of the exponentials is the same, we can equate the exponents: \( p = 5050 \).

Step 5: Simplify the Logarithmic Equation

Now we need to find \( q \) such that \( \ln(q) = \sum_{n=1}^{100} \ln(n) \). Notice that this sum is equivalent to the natural logarithm of the product: \( \ln(q) = \ln(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100) \).

Step 6: Solve for q

The expression \( 1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100 \) is the factorial of 100, written as \( 100! \). Thus, \( q = 100! \).

Key Concepts

Arithmetic Series

An arithmetic series is a way to sum a list of numbers that follow a simple sequence. In such sequences, each term after the first is derived by adding a constant to the previous term. This constant is known as the "common difference."

To find the sum of an arithmetic series, we use the formula:

• The sum of the first \( n \) numbers in the series is \( \frac{n(n+1)}{2} \).

This formula is incredibly efficient, converting our problem into basic multiplication and division.

For example, to find the sum of the numbers from 1 to 100:

• n is 100
• The common difference is 1 (since each number increases by 1)
• Apply the formula: \( \frac{100 imes (100+1)}{2} = 5050 \)

By using this straightforward method, we quickly determine that the sum is 5050.

Understanding this concept aids greatly when solving exponential equations since many times, simplifications can bring the problem down to summing a sequence like this.

Factorial

The concept of a "factorial" applies to non-negative integers. It is the product of all positive integers less than or equal to a specified number. The factorial of a number \( n \) is denoted as \( n! \).

For instance, \( 5! \) ("five factorial") is calculated as:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

In problems involving products over sequences, factorials naturally arise. A classic example is calculating \( 100! \) in expressions like the one we solved for \( q \).

Factorials grow extremely large, extremely fast. Reflect on \( 100! \), which is the product of the numbers from 1 to 100. It's a massive number, often utilized in combinatorics, probability, and calculus. Factorials also prominently feature in simplifying logarithmic equations since:

• \( \ln(n!) = \ln(1 \cdot 2 \cdot 3 \cdot \ldots \cdot n) \)

Natural Logarithms

Natural logarithms (\( \ln \)) are logarithms to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm is extensively used in mathematics, particularly calculus.

Understanding the properties of logarithms can simplify many mathematical problems. Here are a few key properties:

• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln(a^n) = n \ln(a) \)
• \( \ln(1) = 0 \)

In the context of our exercise, applying these properties meant simplifying the sum of logarithms into a single logarithmic expression. For example, the term \( \ln(q) = \ln(1 \cdot 2 \cdot 3 \ldots \cdot 100) \) simplifies due to the factorial representation.

Mastering these properties allows not only simplification of expressions but also enhances problem-solving efficiency across various areas of mathematics.