Table of contents

0. Review of Algebra4h 18m
1. Equations & Inequalities3h 18m
2. Graphs of Equations43m
3. Functions2h 17m
4. Polynomial Functions1h 44m
5. Rational Functions1h 23m
6. Exponential & Logarithmic Functions2h 28m
7. Systems of Equations & Matrices4h 6m
8. Conic Sections2h 23m
9. Sequences, Series, & Induction1h 22m
10. Combinatorics & Probability1h 45m

6. Exponential & Logarithmic Functions

Introduction to Logarithms

College Algebra

6. Exponential & Logarithmic Functions

Introduction to Logarithms

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1:21 minutes

Problem 93

Textbook Question

In Exercises 81–100, evaluate or simplify each expression without using a calculator. e^ln 125

Verified step by step guidance

Recognize that the expression involves the natural exponential function \( e \) and the natural logarithm \( \ln \).

Recall the property of logarithms and exponentials: \( e^{\ln(x)} = x \).

Apply this property to the given expression: \( e^{\ln(125)} \).

Since \( e^{\ln(x)} = x \), substitute \( x \) with 125.

Conclude that \( e^{\ln(125)} = 125 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential and Logarithmic Functions

Exponential functions involve a constant base raised to a variable exponent, while logarithmic functions are the inverse operations of exponentials. The natural logarithm, denoted as ln, uses the base 'e' (approximately 2.718). Understanding the relationship between these functions is crucial for simplifying expressions like e^ln(x).

Properties of Logarithms

Logarithms have specific properties that simplify calculations, such as ln(a^b) = b*ln(a) and e^(ln(a)) = a. These properties allow us to manipulate logarithmic expressions effectively. Recognizing that e and ln are inverse functions helps in evaluating expressions without a calculator.

Evaluating Expressions

Evaluating expressions involves substituting values and simplifying them according to mathematical rules. In this case, understanding how to apply the properties of logarithms and exponentials allows for straightforward simplification of e^ln(125) to yield the final result. Mastery of this concept is essential for solving similar problems.