Problem 275 - Precalculus

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: Recognize the Change-of-Base Formula

The change-of-base formula states that \( \log_{c}(x) = \frac{\log_{k}(x)}{\log_{k}(c)} \). Using this formula, we can change the base of the given logarithm to a more convenient base.

Step 2: Apply Change-of-Base Formula with Base 5

Apply the formula \( \log_{11}(5) = \frac{\log_{5}(5)}{\log_{5}(11)} \). Notice that \( \log_{5}(5) = 1 \) by the property of logarithms, since any number logarithm of itself is 1.

Step 3: Simplify Using Given Values

Substitute \( \log_{5}(11) \) with \( b \). Thus, the expression becomes \( \frac{1}{b} \).

Key Concepts

Change-of-Base Formula

The change-of-base formula is a handy tool in logarithms. It allows us to evaluate logarithms with different bases by converting them into a base we are more comfortable with. Given a logarithm of the form \( \log_{c}(x) \), you can use the formula:

• \( \log_{c}(x) = \frac{\log_{k}(x)}{\log_{k}(c)} \)

Here, \( k \) is a new base of your choosing, typically 10 or \( e \), but you can use any positive number as a base. This transformation works because it balances out the proportion between the original base and the new base, effectively preserving the relationship.
In the given exercise, we used the change-of-base formula to transform \( \log_{11}(5) \) into a fraction involving base 5 logarithms. This step is crucial for applications where you need to express a logarithm in terms of other known logarithmic values, such as \( a = \log_{5}(6) \) and \( b = \log_{5}(11) \).

Properties of Logarithms

Logarithms have several properties that make them especially useful in simplifying and solving equations. Here are some key properties:

• Product Property: \( \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \)
• Quotient Property: \( \log_{b}(\frac{m}{n}) = \log_{b}(m) - \log_{b}(n) \)
• Power Property: \( \log_{b}(m^n) = n \cdot \log_{b}(m) \)

In the context of our exercise, another property comes in handy: \( \log_{b}(b) = 1 \). This stems from the definition of logarithms, where \( b^1 = b \).
We apply this to simplify terms like \( \log_{5}(5) = 1 \), which helped us directly convert the logged value to a fraction form in the change-of-base formula. Understanding these properties allows for breaking down complex logarithmic expressions into manageable parts, paving the way for easier calculations.

Logarithmic Equations

Logarithmic equations involve equations that contain logarithmic expressions. These kinds of equations often require the use of algebraic techniques combined with logarithmic properties to solve them. Here are some steps typically involved in solving them:

• First, simplify the equation using logarithmic identities, such as combining or expanding logs.
• Next, you might need to use the change-of-base formula or convert to an exponential form to isolate the variable.
• Finally, solve for the variable by further simplifying the equation until you reach the solution.

In our exercise, solving the logarithmic equation involved rewriting \( \log_{11}(5) \) in terms of known variables \( a \) and \( b \). By applying our properties and formulas, such as using given information \( \log_{5}(11) = b \), it simplifies to \( \frac{1}{b} \).
Practicing these steps builds a comprehensive understanding and makes solving similar equations in homework or exams much more intuitive.