Chapter 5: Problem 47
Find \(p\) such that \(e^{p}=e \cdot e^{2} \cdot e^{3} \ldots e^{100}\). Find \(q\) so that \(\ln (q)=\sum_{n=1}^{100} \ln (n)\)
Short Answer
Expert verified
\( p = 5050 \), \( q = 100! \).
Step by step solution
01
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: \( e^p = e^{1+2+3+\ldots+100} \).
02
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 \( \frac{100 \cdot (100+1)}{2} \).
03
Compute the Sum
Calculate the value of the sum: \( \frac{100 \cdot 101}{2} = 5050 \). Therefore, \( e^p = e^{5050} \).
04
Solve for p
Since the base of the exponentials is the same, we can equate the exponents: \( p = 5050 \).
05
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: \( \ln(q) = \ln(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100) \).
06
Solve for q
The expression \( 1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100 \) is the factorial of 100, written as \( 100! \). Thus, \( q = 100! \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
For example, to find the sum of the numbers from 1 to 100:
Understanding this concept aids greatly when solving exponential equations since many times, simplifications can bring the problem down to summing a sequence like this.
To find the sum of an arithmetic series, we use the formula:
- The sum of the first \( n \) numbers in the series is \( \frac{n(n+1)}{2} \).
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: \( \frac{100 imes (100+1)}{2} = 5050 \)
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:
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:
For instance, \( 5! \) ("five factorial") is calculated as:
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
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:
Mastering these properties allows not only simplification of expressions but also enhances problem-solving efficiency across various areas of mathematics.
Understanding the properties of logarithms can simplify many mathematical problems. Here are a few key properties:
- \( \ln(ab) = \ln(a) + \ln(b) \)
- \( \ln(a^n) = n \ln(a) \)
- \( \ln(1) = 0 \)
Mastering these properties allows not only simplification of expressions but also enhances problem-solving efficiency across various areas of mathematics.