Problem 28 - 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. \(\log _{6}(55)\)

Step-by-Step Solution

Answer

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

Step 1: Change the Base

Use the change-of-base formula to rewrite \(\log_{6}(55)\). According to this formula, \(\log_{b}(x) = \frac{\log_{c}(x)}{\log_{c}(b)}\). Choosing base \(5\), we have: \[ \log_{6}(55) = \frac{\log_{5}(55)}{\log_{5}(6)} \]

Step 2: Express in Terms of Known Values

We will now express \(\log_{5}(55)\). Use the property \(\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)\) to rewrite the expression. Since \(55 = 5 \times 11\), apply the property: \[\log_{5}(55) = \log_{5}(5 \times 11) = \log_{5}(5) + \log_{5}(11) \]

Step 3: Substitute Known Log Values

We know that \(\log_{5}(5) = 1\) because any log of the base itself is always 1. We also have \(\log_{5}(11) = b\). Substitute these values into the equation: \[ \log_{5}(55) = 1 + b \]

Step 4: Combine and Simplify

Now, substitute the value of \(\log_{5}(55)\) into the original change-of-base equation: \[ \log_{6}(55) = \frac{1 + b}{a} \] Thus, the expression for \(\log_{6}(55)\) in terms of \(a\) and \(b\) is \(\frac{1 + b}{a}\).

Key Concepts

Properties of Logarithms

Logarithms have a set of properties that allow us to manipulate and simplify logarithmic expressions effectively. One fundamental property is the product rule, which states that for any positive numbers, the logarithm of a product is equal to the sum of the logarithms. In mathematical terms, this is written as:

• \(\log_b(xy) = \log_b(x) + \log_b(y)\)

Another key property is the power rule, helping to manage exponents within logs. It explains that the logarithm of a number raised to an exponent is equivalent to the exponent times the logarithm of the base number:

• \(\log_b(x^n) = n \cdot \log_b(x)\)

Lastly, the quotient rule is beneficial when dealing with division inside a logarithm. It can be stated as:

• \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)

These properties not only simplify complex expressions but also provide a pathway to expressing unknown log values in terms of known ones.

Logarithmic Expressions

When working with logarithmic expressions, the goal is often to simplify or rewrite the expression using known values or simpler forms. For instance, we sometimes need to express a logarithmic expression like \(\log_6(55)\) in terms of other logarithms that are defined.
In our solution, we utilized properties of logarithms to break down \(\log_5(55)\) into \(\log_5(5 \times 11)\), leveraging the product rule:

• \(\log_5(55) = \log_5(5) + \log_5(11)\)

Knowing that \(\log_5(5) = 1\) because the log of a base to itself is always 1, simplifies our expression as:

• \(\log_5(55) = 1 + b\)

This breakdown is useful as it converts a less familiar log into terms associated with known quantities \(a\) and \(b\), making calculations and substitutions much more straightforward.

Base Conversions

Converting between different bases is a common task when dealing with logarithms. Sometimes, you need to evaluate a log expression in one base from another using the change of base formula. This formula is:

• \(\log_b(x) = \frac{\log_c(x)}{\log_c(b)}\)

This conversion is crucial when you have logs with bases that aren't readily available on calculators or need to express them in terms of log values with known bases. In our solution, we used base 5, converting the original expression \(\log_6(55)\) to base 5 to use the known values of \(a = \log_5(6)\) and \(b = \log_5(11)\).
Through this method, we expressed \(\log_6(55)\) in terms of \(\frac{1 + b}{a}\) by substituting the known values and simplifying. This approach of base conversion not only helps in solving problems but also deepens the understanding of how logarithms function across different numerical bases.