Problem 276 - 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. $$ \log _{6}(55) $$

Step-by-Step Solution

Answer

\( \frac{1+b}{a} \)

Step 1: Change of Base Formula

To rewrite \( \log_{6}(55) \), we first use the change of base formula, which states: \( \log_{b}(x) = \frac{\log_{c}(x)}{\log_{c}(b)} \). We can choose 5 as the new base to utilize the given logs. Thus, \( \log_{6}(55) = \frac{\log_{5}(55)}{\log_{5}(6)} \).

Step 2: Rewrite Logarithms in the Numerator and Denominator

We already know \( \log_{5}(6) = a \). We need to express \( \log_{5}(55) \) using properties of logarithms. Notice that 55 can be rewritten as 5 * 11, so \( \log_{5}(55) = \log_{5}(5 \times 11) \). Using the property \( \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \), we get: \( \log_{5}(55) = \log_{5}(5) + \log_{5}(11) \).

Step 3: Substitute Known Values

Now substitute: \( \log_{5}(5) = 1 \) (since any log where the base and the argument are the same is 1), and we know from the problem that \( \log_{5}(11) = b \). Hence, \( \log_{5}(55) = 1 + b \).

Step 4: Final Expression

Substitute back into the expression from Step 1: \( \log_{6}(55) = \frac{1 + b}{a} \). Thus, the original expression is rewritten in terms of \(a\) and \(b\).

Key Concepts

Properties of Logarithms

Logarithms have special properties that help simplify expressions. Understanding these properties can make calculations easier and more intuitive. Here are some crucial properties of logarithms:

• Product Property: The logarithm of a product is the sum of the logarithms. Mathematically, this property is expressed as \( \log_b(mn) = \log_b(m) + \log_b(n) \).
• Quotient Property: The logarithm of a quotient is the difference between the logarithms. This is stated as \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \).
• Power Property: The logarithm of a power is the exponent times the logarithm of the base. Formally written as \( \log_b(m^n) = n \cdot \log_b(m) \).

These properties allow us to break down complicated logarithmic expressions into manageable parts.
For example, in the original problem, we use the product property to simplify \( \log_5(55) \) by recognizing it as \( 5 \times 11 \).

Logarithmic Expressions

Logarithms help us express large numbers compactly. They are handy in various fields such as science, engineering, and computer science. Logarithms convert multiplication into addition, division into subtraction, and exponents into multiplication.
This transformation is particularly useful when working with very large or small numbers.

• Basic Definition: A logarithm \( \log_b(x) \) answers the question: "To what power must the base \( b \) be raised to obtain \( x \)?"
• Change of Base Formula: This formula is essential for rewriting logarithmic expressions with a different base. It is given by \( \log_b(x) = \frac{\log_c(x)}{\log_c(b)} \), allowing you to pick any base \( c \), often one that simplifies your problem.

The change of base formula was critical in the given exercise, transforming \( \log_6(55) \) to a base 5 expression, where calculations are easier given the known logarithms \( \log_5(6) \) and \( \log_5(11) \).

Rewrite Expressions in Terms of Variables

Rewriting expressions in terms of variables is a powerful technique, especially in algebra and calculus. When you rewrite expressions, you simplify problems and make them more intuitive.

• In the exercise, you are tasked with rewriting \( \log_6(55) \) in terms of variables \( a \) and \( b \).
• Given: \( \log_5(6) = a \) and \( \log_5(11) = b \), the goal is to express everything using these known values.
• This involves first changing the base of the logarithm using the change of base formula. Then, using the properties of logarithms, you rewrite the expression to align with the given variables.

Once you substitute known results, you achieve a final expression in a simple form, like \( \frac{1+b}{a} \), making it easier to understand and work with.
This method teaches you how to manipulate variables and expressions creatively, an essential skill in advanced mathematics. Learning to express complex ideas using simple variables not only simplifies the problem but also broadens your problem-solving toolkit.