Problem 35 - Calculus and its applications

Question

Given \(\ln 4=1.3863\) and \(\ln 5=1.6094,\) find each value. Do not use a calculator. $$ \ln (4 e) $$

Step-by-Step Solution

Answer

\( \ln(4e) = 2.3863 \).

Step 1: Apply the Logarithmic Identity

To find \( \ln(4e) \), we start by using the logarithmic identity \( \ln(ab) = \ln a + \ln b \). Here, \( a = 4 \) and \( b = e \), so we have \( \ln(4e) = \ln 4 + \ln e \).

Step 2: Evaluate \( \ln e \)

Recall that the natural logarithm of \( e \), \( \ln e \), is defined to be 1. Therefore, \( \ln e = 1 \).

Step 3: Substitute Known Values

Substitute the known values into the equation from Step 1: \( \ln(4e) = \ln 4 + 1 \). Since \( \ln 4 = 1.3863 \) (provided in the problem), we substitute to get \( \ln(4e) = 1.3863 + 1 \).

Step 4: Perform the Addition

Add the two numbers: \( 1.3863 + 1 = 2.3863 \).

Key Concepts

Logarithmic Identities

Logarithmic identities are powerful tools in mathematics that help simplify complex logarithmic expressions. One key identity is the product rule for logarithms, which states:

• For any positive numbers \( a \) and \( b \), \( \ln(ab) = \ln a + \ln b \).

This identity allows you to break down the logarithm of a product into the sum of the logarithms, making calculations more manageable.
In the given exercise, we see this identity in action with the expression \( \ln(4e) \). By identifying \( a = 4 \) and \( b = e \), the expression splits into \( \ln 4 + \ln e \). The product rule is especially handy when you know the logarithms of individual numbers involved, as demonstrated by the provided values \( \ln 4 = 1.3863 \) and \( \ln e = 1 \). This identity can also be adapted for different logarithmic bases, applying similarly for common logarithms (base 10) as well. Understanding and mastering these identities is crucial, as they simplify many algebraic and calculus problems.

Properties of Logarithms

The properties of logarithms are essential for performing calculations without a calculator. These properties exploit the relationships between operations to aid in simplifying and solving logarithmic expressions. Among these properties are:

• Product Property: \( \ln(ab) = \ln a + \ln b \)
• Quotient Property: \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \)
• Power Property: \( \ln(a^b) = b \times \ln a \)

Reviewing these properties simplifies the task of manipulating logarithmic terms. For instance, the product property was used to break down \( \ln(4e) \) into \( \ln 4 + \ln e \), simplifying the evaluation process. Recognizing these relationships helps you perform accurate calculations and solve complex logarithmic problems by breaking them into simpler parts.
Additionally, being familiar with these properties provides the groundwork for further mathematical concepts, such as solving logarithmic equations and integrating functions in calculus. They offer a powerful toolkit for students navigating both theoretical and applied mathematics.

Evaluation of Logarithmic Expressions

Evaluating logarithmic expressions involves substituting specific values and applying logarithmic identities to simplify the expressions. Let's consider our example of evaluating \( \ln(4e) \). We used both the identity \( \ln(ab) = \ln a + \ln b \) and the known value \( \ln e = 1 \) to simplify the expression.
Here's a step-by-step breakdown of the evaluation process:

• Identify Known Values: Start with what is given. In this exercise, \( \ln 4 = 1.3863 \) and \( \ln e = 1 \) are provided, allowing for immediate substitution.
• Apply Logarithmic Identity: Use identities like the product rule to simplify using the expression \( \ln(4e) = \ln 4 + \ln e \).
• Perform Substitution and Simplification: Replace \( \ln 4 \) and \( \ln e \) with their numerical values, resulting in \( \ln(4e) = 1.3863 + 1 \).
• Final Calculation: Add the numbers to complete the evaluation, resulting in \( 2.3863 \).

This process not only demystifies the calculation but also reinforces the logical flow from knowing properties to applying calculations effectively. Mastery of evaluating logarithmic expressions equips students to handle real-world problems and advance in mathematical literacy.