Problem 45 - Intermediate Algebra

Question

Evaluate each expression without using a calculator. $$ \ln \frac{1}{e} $$

Step-by-Step Solution

Answer

The value of \( \ln \frac{1}{e} \) is \( -1 \).

Step 1: Understand the Problem

We need to evaluate the natural logarithm of \( \frac{1}{e} \). Recall that the natural logarithm, \( \ln \), is the inverse operation to exponentiation with base \( e \).

Step 2: Apply the Logarithm Property

Use the property of logarithms that states \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \). Here, \( a = 1 \) and \( b = e \), so this becomes \( \ln(1) - \ln(e) \).

Step 3: Evaluate Logarithm of One

Recall that \( \ln(1) = 0 \) because \( e^0 = 1 \). Substitute this into our expression to get \( 0 - \ln(e) \).

Step 4: Evaluate Logarithm of e

We know that \( \ln(e) = 1 \) because \( e^1 = e \). Substitute this into the expression to get \( 0 - 1 \).

Step 5: Simplify the Expression

Perform the subtraction: \( 0 - 1 = -1 \). Therefore, \( \ln \frac{1}{e} = -1 \).

Key Concepts

Logarithm Properties

Logarithms have unique properties that make them useful for simplifying expressions and solving equations. One fundamental property is the logarithm of a quotient: \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \). This property indicates that the logarithm of a division can be expressed as the difference of two logarithms.
This is particularly helpful when evaluating expressions involving fractions.
Another key property is \( \ln(1) = 0 \). This is because the exponent of 0 will always yield 1 in any base. It is crucial to remember these properties as they can transform complex logarithmic problems into much simpler arithmetic operations.

Inverse Operations

The concept of inverse operations is central to understanding logarithms. In mathematics, an inverse operation reverses the effect of another operation.
Specifically, a logarithm is the inverse of exponentiation. When dealing with natural logarithms, the base of the exponentiation is \( e \). This means that \( \ln(e^x) = x \) and \( e^{\ln(x)} = x \).
These equations illustrate how logarithms "undo" exponentiation. If you understand one process, the other becomes more intuitive. Embracing this concept allows you to navigate operations between logarithms and exponentials seamlessly.

Exponentiation

Exponentiation is a mathematical operation involving two numbers: a base and an exponent. The base is raised to the power of the exponent, such as \( e^x \). Another critical piece of knowledge is knowing \( e^0 = 1 \) and \( e^1 = e \).
These basics help in evaluating natural logarithms. The understanding that \( e \, \text{and}\, \ln \) are inverse operations at different stages is vital. This relationship allows us to simplify expressions such as \( \ln(e) = 1 \), which often appear in mathematical exercises, reducing complexity significantly.

Evaluate Expressions

Evaluating expressions often involves breaking down the problem step by step. For a logarithm like \( \ln \frac{1}{e} \), apply the properties and relationships you know. Start with the recognition that this is equivalent to \( \ln(1) - \ln(e) \).
Recall the specific logarithm values: \( \ln(1) = 0 \) and \( \ln(e) = 1 \).
These values emerge from the very definition of logarithms and their inverse relationships. Consequently, the expression simplifies to \( 0 - 1 = -1 \).

• Recognize essential logarithmic identities.
• Simultaneously use arithmetic and benchmark knowledge (like easily remembered logarithm values).

This systematic approach helps simplify and evaluate expressions effectively without a calculator.