Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, suppose \(\log _{5}(6)=a\) and \(\log _{5}(11)=b .\) Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of \(a\) and \(b .\) Show the steps for solving. \(\log _{11}(5)\)

Step-by-Step Solution

Answer

\( \log_{11}(5) = \frac{1}{b} \)

Step 1: Apply Change of Base Formula

The change of base formula for logarithms is given by \( \log_{b}(x) = \frac{\log_{c}(x)}{\log_{c}(b)} \) for any positive number \( c \). We'll apply this formula here with base 5: \[ \log_{11}(5) = \frac{\log_{5}(5)}{\log_{5}(11)} \] Given \( \log_{5}(11) = b \), this expression simplifies to: \[ \log_{11}(5) = \frac{1}{b} \] since \( \log_{5}(5) = 1 \).

Step 2: Conclusion: Final Expression

Using logarithmic identities and the given values: \( \log_{11}(5) = \frac{1}{b} \). This expression is rewritten in terms of \( a \) and \( b \) as \( \frac{1}{b} \), where \( b = \log_{5}(11) \).

Key Concepts

Logarithm Properties

Logarithms are fascinating mathematical tools that help us understand how to scale numbers and manage exponential growth or decay. There are several properties of logarithms that make them very useful in mathematics:

• Product Property: The logarithm of a product is the sum of the logarithms of the individual factors, i.e., \( \log_b(xy) = \log_b(x) + \log_b(y) \).


• Quotient Property: The logarithm of a quotient is the difference of the logarithms of the numerator and the denominator, i.e., \( \log_b\left( \frac{x}{y} \right) = \log_b(x) - \log_b(y) \).


• Power Property: The logarithm of an exponentiated value is the exponent times the logarithm of the base, i.e., \( \log_b(x^y) = y \cdot \log_b(x) \).

Understanding these properties is the key to solving various logarithmic equations and can simplify calculations significantly. These properties allow us to break down complex expressions into simpler parts, which is crucial for solving real-world problems efficiently.

Logarithmic Identities

Logarithmic identities are special equalities that hold true for all values that satisfy the conditions within logarithmic expressions. These identities are derived from the basic properties of logarithms and help in solving equations and rewriting expressions.

• Identity Property: Any nonzero number's logarithm to its own base is 1, such as \( \log_b(b) = 1 \).


• Zero Property: The logarithm of 1 in any base is 0, for instance, \( \log_b(1) = 0 \).


• Change of Base Formula: This identity allows the conversion between various logarithmic bases and is crucial when the calculator only supports a specific base: \( \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \) for any positive and different bases \( k \).

Mastering these identities enables us to transition between different base systems, significantly broadening our ability to handle various mathematical and applied problems.

Base Conversion

Base conversion in logarithms refers to the ability to switch from one logarithmic base to another. This process is essential when we want to use a base not directly supported by our tools or when simplifying calculations.The most common tool for base conversion is the Change of Base Formula:\[ \log_b(x) = \frac{\log_c(x)}{\log_c(b)} \]This formula allows us to calculate the logarithm of any number \( x \) to any base \( b \) by using a new base \( c \). A typical choice for \( c \) is either 10 (common logarithm) or \( e \) (natural logarithm), as these are the bases commonly available on calculators.
For example, given \( \log_5(6) = a \) and \( \log_5(11) = b \), you can convert \( \log_{11}(5) \) as shown:

• Apply the formula: \( \log_{11}(5) = \frac{\log_5(5)}{\log_5(11)} \)
• Use \( \log_5(5) = 1 \) due to the identity property, and simplify to \( \frac{1}{b} \).

By applying this formula, you can change the base of any logarithmic expression, facilitating the solving of complex logarithmic equations.

Step-by-Step Problem Solving

Solving logarithmic problems step by step is a methodological approach to breaking down complex problems into manageable parts. Here's how you can approach these problems efficiently:

• Understand the Problem: Read the problem carefully, identify what is given (e.g., \( \log_5(6) = a \), \( \log_5(11) = b \)) and what needs to be found (e.g., \( \log_{11}(5) \)).


• Choose a Strategy: Decide which logarithmic property or identity can be utilized to simplify or solve the problem. Here, the change of base formula is applied.


• Apply the Formula: Substitute the known values into the formula. For the example problem, using \( \log_{11}(5) = \frac{1}{b} \), simplifies the task by converting the base.


• Verify Your Solution: Always double-check your steps and final answer for accuracy, ensuring consistent results for the values of \( a \) and \( b \).

By following these steps, you're not just solving the problem at hand but also gaining a deeper insight into how logarithmic expressions work, which will assist you in tackling more complex problems in the future.