Problem 29 - 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. \(\log _{11}\left(\frac{6}{11}\right)\)

Step-by-Step Solution

Answer

\( \log_{11}\left(\frac{6}{11}\right) = \frac{a - b}{b} \).

Step 1: Recognize the problem and relevant formulas

The task is to express \( \log_{11}\left(\frac{6}{11}\right) \) in terms of \( a \) and \( b \). We will use the change-of-base formula for logarithms, which states \( \log_{c}(x) = \frac{\log_{b}(x)}{\log_{b}(c)} \).

Step 2: Apply the change-of-base formula

Apply the change-of-base formula: \( \log_{11}\left(\frac{6}{11}\right) = \frac{\log_{5}\left(\frac{6}{11}\right)}{\log_{5}(11)} \). This reformulates the logarithm in terms of the base 5, using our given expressions.

Step 3: Use the properties of logarithms

According to the properties of logarithms, \( \log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y) \). Thus, we have \( \log_{5}\left(\frac{6}{11}\right) = \log_{5}(6) - \log_{5}(11) \).

Step 4: Substitute known values

We know that \( \log_{5}(6) = a \) and \( \log_{5}(11) = b \). Substitute these into the equation from Step 3: \( \log_{5}\left(\frac{6}{11}\right) = a - b \).

Step 5: Complete substitution in initial expression

Substitute \( a - b \) from Step 4 into the expression obtained in Step 2: \( \frac{a - b}{b} \). Thus, \( \log_{11}\left( \frac{6}{11} \right) = \frac{a - b}{b} \).

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools that help simplify complex expressions involving logarithms. A core property is the quotient rule, which states that the logarithm of a division is the difference of the logarithms. This can be expressed as \[\log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y)\].
This property is used to break down more complicated logarithmic expressions into simpler components. Ul>

When given a division inside a logarithm, you can split it into two separate log terms.

This allows you to work with each part of the expression separately.

Using these properties allows mathematicians to solve equations more efficiently by reducing complex expressions into manageable parts.

Logarithm Base Conversion

Logarithm base conversion is a crucial skill when dealing with logarithms, especially when you need to express a logarithm from one base to another. The change-of-base formula is your primary tool here: \[\log_{c}(x) = \frac{\log_{b}(x)}{\log_{b}(c)}\].
This formula allows you to convert a logarithm from any base to a common base, which simplifies calculations and comparisons.

• Using a common base helps in combining or comparing different logarithms.
• For calculations, converting logarithms to a base that you're more familiar with (like base 10 or base 5) can make the process easier.

In our example, we use this formula to change \(\log_{11}\) into an expression involving \(\log_{5}\) so that known values of \(\log_{5}(6) = a\) and \(\log_{5}(11) = b\) can be used.

Algebraic Manipulation

Algebraic manipulation plays a significant role in rewriting logarithmic expressions in a simpler form. By using known values and properties of logarithms, you can substitute and rearrange expressions to isolate variables or express them in a desired form.
This technique often involves several steps:

• Use the properties of logarithms to simplify expressions (like turning a division inside a log into a subtraction of two logs).
• Substitute known quantities or expressions to further simplify the equation.
• Solve for the desired variable or form based on the given conditions.

In our exercise, after using the logarithm properties and change-of-base formula, we substituted known values \(a\) and \(b\) to eventually express \(\log_{11}\left(\frac{6}{11}\right)\) in simpler terms of \(\frac{a-b}{b}\). This stepwise manipulation is essential in algebra to manage and solve logarithmic equations effectively.