Problem 40 - Intermediate Algebra

Question

Write each logarithm as a difference. Then simplify, if possible. See Example 3 . $$ \ln \frac{27}{e} $$

Step-by-Step Solution

Answer

The expression simplifies to \(\ln(27) - 1\).

Step 1: Understand the Logarithmic Property

We will use the property of logarithms which states that for any positive numbers \(a\) and \(b\), \(\ln \left(\frac{a}{b}\right) = \ln(a) - \ln(b)\). Our goal is to express the given logarithm as a difference of two logarithms.

Step 2: Apply the Property to the Given Logarithm

Apply the property from Step 1 to the given logarithm \(\ln \frac{27}{e}\). This can be written as a difference: \(\ln(27) - \ln(e)\).

Step 3: Recognize the Simplification Possibility

Recall that \(\ln(e) = 1\) since the natural logarithm of the base \(e\) is 1. Thus, \(\ln(27) - \ln(e)\) simplifies to \(\ln(27) - 1\).

Key Concepts

Understanding Natural Logarithms

Natural logarithms are based on the constant \( e \), which is approximately equal to 2.718. This constant is fundamental in mathematics, commonly used in the study of exponential growth and decay. Natural logarithms help us solve problems where the base of the logarithm is \( e \). This is why natural logarithms are often written as \( \ln \) rather than using a subscript to indicate the base, like \( \log_e \).

The key property of natural logarithms is the ability to simplify expressions involving \( e \). For example, \( \ln(e) = 1 \), because the logarithm essentially asks the question: "What power should \( e \) be raised to in order to get \( e \)?" Since \(e^1 = e\), the answer is 1. This property makes computations involving \( e \) straightforward and easy to manage in logarithmic expressions.

Simplification of Logarithms

Simplifying logarithms involves using properties that transform expressions into simpler or alternative forms. One of the most useful properties involves quotients, such as the one used in our exercise. With this property, \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), we can break down a complex logarithm into a more manageable expression.

This simplification process is helpful for several reasons:

• It makes equations easier to solve by breaking them into component parts.
• It reveals hidden connections between numbers by expressing them as sums or differences.
• It allows for greater flexibility in mathematical manipulation and aids in solving equations.

In our exercise, \( \ln\left(\frac{27}{e}\right) \) is simplified using this property, transforming it into \( \ln(27) - \ln(e) \), which further simplifies using the property that \( \ln(e) = 1 \) to \( \ln(27) - 1 \). This simplification process is at the heart of working efficiently with logarithmic calculations.

Difference of Logarithms

The difference of logarithms is a direct consequence of the division property of logarithms. It provides an intuitive way to understand how logarithms distribute over division. In our example, we used this property to express \( \ln\left(\frac{27}{e}\right) \) as a difference, \( \ln(27) - \ln(e) \).

To grasp this concept:

• Consider the properties of the functions you are working with. Logarithms turn division into subtraction, making complex expressions easier to handle.
• Remember that the property is \( \ln(a) - \ln(b) \) equals \( \ln\left(\frac{a}{b}\right) \). This can simplify calculations and is especially useful when dealing with ratios or fractions.
• Recognizing and applying the difference of logarithms can streamline the problem-solving process, reduce computational overhead, and clarify the relationship between numbers in logarithmic form.

So, when you see a complex fraction within a logarithm, remember that it can often be expressed as a difference, simplifying the expression and shedding light on the problem at hand.