Problem 41 - College Algebra

Question

Prove that \(\log _{b}(n)=\frac{1}{\log _{n}(b)}\) for any positive integers \(b>1\) and \(n>1 .\)

Step-by-Step Solution

Answer

The relationship \(\log_{b}(n) = \frac{1}{\log_{n}(b)}\) holds due to the properties of logarithms.

Step 1: Understand the Logarithmic Relationship

To prove that \(\log_{b}(n)=\frac{1}{\log_{n}(b)}\), we need to understand what each side of the equation represents. The expression \(\log_{b}(n)\) is the power to which we must raise \(b\) to obtain \(n\). In other words, if \(\log_{b}(n) = x\), then \(b^x = n\). Similarly, \(\log_{n}(b)\) is the power to which we must raise \(n\) to obtain \(b\). If \(\log_{n}(b) = y\), then \(n^y = b\).

Step 2: Set Up the Equations

Using the definitions from Step 1, we set up the two equations:1. \(b^x = n\) where \(x = \log_{b}(n)\).2. \(n^y = b\) where \(y = \log_{n}(b)\).

Step 3: Solve for x and y

From the equation \(b^x = n\), taking the logarithm (base 10 or e, consistently) on both sides gives us:\[ x = \frac{\log(n)}{\log(b)} \]Similarly, for \(n^y = b\), taking the logarithm gives:\[ y = \frac{\log(b)}{\log(n)} \]

Step 4: Relationship Between x and y

Now, observe the relationship between \(x\) and \(y\):\[ x = \frac{\log(n)}{\log(b)} = \frac{1}{\frac{\log(b)}{\log(n)}} = \frac{1}{y} \]Thus, \(x = \frac{1}{y}\), which means \(\log_{b}(n) = \frac{1}{\log_{n}(b)}\).

Step 5: Conclusion

We have shown through the manipulation of logarithmic identities that indeed, \(\log_{b}(n) = \frac{1}{\log_{n}(b)}\). This relationship is a consequence of the properties of logarithms and the definition itself.

Key Concepts

Logarithmic Relationships

When working with logarithms, it's important to understand the relationships they express. Logarithms help us to comprehend exponential growth and decay by transforming multiplicative processes into additive ones. In simpler terms, logarithms answer the question: "To what power must we raise a certain base to obtain a particular number?" For example, in the equation \( \log_b(n) \), we find out what power we need to raise \( b \) to get \( n \). These relationships allow us to solve exponential equations more easily by breaking them down into manageable parts. Understanding these foundational relationships is crucial for grasping more complex logarithmic concepts and equations like the one presented in this exercise.

Properties of Logarithms

Logarithms have several key properties that often come in handy in mathematical proofs and problem solving. Here are some important properties:

• Product Property: \( \log_b(mn) = \log_b(m) + \log_b(n) \)
• Quotient Property: \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \)
• Power Property: \( \log_b(m^p) = p \cdot \log_b(m) \)
• Change of Base Formula: \( \log_b(n) = \frac{\log_k(n)}{\log_k(b)} \) for any base \( k \)

These properties simplify complex logarithmic expressions and help in solving logarithmic equations. In our exercise, we used the change of base formula extensively to demonstrate the equation \( \log_b(n) = \frac{1}{\log_n(b)} \). By focusing on the change of base property, we were able to manipulate and establish the proof for this logarithmic identity.

Mathematical Proof

A mathematical proof is a process of establishing the truth of a given statement based on logical reasoning and established facts. In this exercise, we proved the logarithmic identity \( \log_b(n) = \frac{1}{\log_n(b)} \) using known properties of logarithms. To start, we used the definitions and relationships that describe what each logarithmic expression means. By setting \( b^x = n \) (where \( x = \log_b(n) \)) and \( n^y = b \) (where \( y = \log_n(b) \)), we created a framework to demonstrate the connection between the two logarithmic forms. Next, by applying the natural or common logarithms on both sides of these equations, we derived expressions for \( x \) and \( y \), establishing that \( x = \frac{1}{y} \). This step-by-step logical reasoning confirmed the identity, showcasing how mathematical proofs not only validate concepts but also deepen our understanding of the structure of numbers.

Educational Algebra

Educational algebra involves teaching and learning the key concepts and operations of algebraic expressions, equations, and functions. A critical part of algebraic education is understanding how to manipulate and solve different types of equations, such as exponential and logarithmic equations. This particular exercise helps students to understand logarithmic relationships and properties through hands-on experience with algebraic manipulation and proof strategies. Here are a few tips when tackling similar algebraic problems:

• Begin by understanding the problem: Identify what is given and what needs to be proved.
• Make use of algebraic identities and properties: These are your tools to simplify and solve problems.
• Practice logical reasoning: This is key to forming coherent mathematical proofs.
• Tackle problems step-by-step: Break down complex problems into smaller, manageable parts.

By focusing on these aspects, students can strengthen their problem-solving skills and build a solid foundation in algebra, allowing them to better tackle various mathematical challenges.