Problem 28 - College Algebra

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. $$ \log _{6}(55) $$

Step-by-Step Solution

Answer

\( \log_{6}(55) = \frac{1 + b}{a} \).

Step 1: Apply the Change of Base Formula

Use the change of base formula to rewrite the expression \( \log_{6}(55) \). The change of base formula states that \( \log_{b}(x) = \frac{\log_{k}(x)}{\log_{k}(b)} \), for any positive \( k eq 1 \). Choose base 5 to relate it to known terms: \( \log_{6}(55) = \frac{\log_{5}(55)}{\log_{5}(6)} \).

Step 2: Express \(\log_{5}(55)\) Using Logarithm Properties

Recognize \( 55 \) as the product of \( 5 \) and \( 11 \): \( 55 = 5 \times 11 \). Use the property \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \) to write \( \log_{5}(55) = \log_{5}(5 \times 11) = \log_{5}(5) + \log_{5}(11) \). Since \( \log_{5}(5) = 1 \) and \( \log_{5}(11) = b \), it becomes \( 1 + b \).

Step 3: Substitute Known Values and Simplify

Substitute the simplified \( \log_{5}(55) \) expression into the fraction from Step 1: \( \log_{6}(55) = \frac{1 + b}{\log_{5}(6)} \). We know \( \log_{5}(6) = a \), so substitute \( a \) for \( \log_{5}(6) \) to obtain \( \log_{6}(55) = \frac{1 + b}{a} \).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms can simplify complex expressions. They help in breaking down intricate calculations into manageable parts.
One essential property is the **product rule**: \(\log_b(xy) = \log_b(x) + \log_b(y)\). This property allows us to split a logarithm of a product into a sum of logarithms, which is extremely helpful when dealing with large numbers. For example, for \(55 = 5 \times 11\),

• \(\log_{5}(55) = \log_{5}(5 \times 11)\)
• This becomes \(\log_{5}(5) + \log_{5}(11)\)

Another relevant property is the **identity property**: \(\log_b(b) = 1\). This is simple yet powerful because it helps to easily discard terms when the base matches the argument.
Together, these properties assist in converting complex logarithms into simpler forms, making manipulation and computation easier.

Logarithm Expressions

Logarithm expressions might seem intimidating at first glance, but they boil down to simple, repetitive patterns using their properties.
A logarithm evaluates how many times one number (the base) must be multiplied to get another number. Logarithm expressions can initially be complex, but you can simplify them by expressing their arguments in terms of recognizable factors or sums.
In the example \(\log_5(55)\), we use the change of base formula to rewrite \(\log_6(55)\) in terms of log base 5. This helps to leverage known relationships like \(\log_{5}(6) = a\) and \(\log_{5}(11) = b\). Each expression can then be broken down, such as understanding \(55\) as \(5 \times 11\), making it easier to apply the logarithm properties effectively.

Algebraic Manipulation

Algebraic manipulation is an invaluable skill when working with logarithmic equations, enabling us to rearrange and simplify expressions easily.
The core idea is using known values to substitute and simplify logarithmic expressions. For instance, after expanding \(\log_{5}(55)\) to \(1 + b\) using logarithmic properties, we insert this expression into another formula, such as the change of base output. The task may involve further manipulation of the expression \(\frac{1 + b}{a}\) to reach a final form.

• This involves substituting known values like \(\log_{5}(6) = a\).
• Finally, this replaces the initial expression with known terms: \(\log_{6}(55) = \frac{1 + b}{a}\).

By mastering algebraic manipulation, we gain the power to simplify these seemingly complex expressions efficiently, using substitution and known identities to our advantage.