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

Step-by-Step Solution

Answer

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

Step 1: Apply Change of Base Formula

Use the change of base formula to express the given logarithm with base 11 in terms of base 5. The change of base formula states \( \log_{b}(x) = \frac{\log_{c}(x)}{\log_{c}(b)} \). Here, apply it to \( \log_{11}(\frac{6}{11}) \) to rewrite it in terms of base 5: \[ \log_{11} \left( \frac{6}{11} \right) = \frac{\log_{5} \left( \frac{6}{11} \right)}{\log_{5}(11)} \].

Step 2: Apply Logarithm of a Quotient

Use the logarithm of a quotient property \( \log_{c}(\frac{x}{y}) = \log_{c}(x) - \log_{c}(y) \) on the numerator: \( \log_{5} \left( \frac{6}{11} \right) = \log_{5}(6) - \log_{5}(11) \).

Step 3: Substitute Known Values

Substitute the known values \( \log_{5}(6) = a \) and \( \log_{5}(11) = b \) into the expression derived from Step 2. Thus, \( \log_{5} \left( \frac{6}{11} \right) = a - b \).

Step 4: Complete Expression

Substitute the expression from Step 3 into Step 1. Therefore, \[ \log_{11} \left( \frac{6}{11} \right) = \frac{a - b}{b} \].

Key Concepts

Change of Base Formula

The change of base formula is an essential tool in dealing with logarithmic expressions of different bases. It allows us to convert a logarithm with any base to a different base that might be more convenient for calculations. This is particularly useful when the base of the logarithm isn't readily programmable or a calculator's default. The formula is expressed as:
\[ \log_{b}(x) = \frac{\log_{c}(x)}{\log_{c}(b)} \]
where:

• \( b \) is the original base,
• \( x \) is the argument of the logarithm,
• and \( c \) is the new base we are converting to.

In our exercise, converting \( \log_{11}(\frac{6}{11}) \) to base 5, makes use of this formula to simplify the expression further using the given values \( \log_{5}(6) = a \) and \( \log_{5}(11) = b \). By changing the base from 11 to 5, we obtain a quotient of logs that solely relies on base 5.

Properties of Logarithms

Understanding the properties of logarithms is fundamental in manipulating and simplifying complex logarithmic expressions. Some essential properties include:

• Product Property: \( \log_{c}(xy) = \log_{c}(x) + \log_{c}(y) \)
• Quotient Property: \( \log_{c}(\frac{x}{y}) = \log_{c}(x) - \log_{c}(y) \)
• Power Property: \( \log_{c}(x^p) = p \cdot \log_{c}(x) \)

In our problem, the quotient property is particularly helpful. It assists in breaking down a divided argument into a simple difference of two logs. Applying these rules effectively helps in simplifying our expression from a single complex logarithm to a series of operations that use known values.

Logarithm of a Quotient

The logarithm of a quotient property plays a critical role in simplifying logarithmic expressions that include fractions as their arguments. According to this property:
\[ \log_{c}(\frac{x}{y}) = \log_{c}(x) - \log_{c}(y) \]
This property turns the division inside the log into a subtraction. In the case of \( \log_{5}(\frac{6}{11}) \), the expression turns into \( \log_{5}(6) - \log_{5}(11) \). This form is particularly useful as it allows us to substitute known values for \( \log_{5}(6) \) and \( \log_{5}(11) \), noted as \( a \) and \( b \) respectively, into the expression. This simple transformation provides a cleaner and more manageable form to work with.

Rewriting Expressions

Rewriting expressions is a powerful technique in mathematics to find simpler equivalent forms of complex expressions. In logarithms, this often involves applying properties of logarithms or using an alternative base to streamline expressions for clearer understanding or easier computation. For example, in our exercise:

• We first use the change of base formula to rewrite \( \log_{11}(\frac{6}{11}) \) in terms of base 5 logs.
• Then, we leverage the quotient property to express it as \( \log_{5}(6) - \log_{5}(11) \).

Finally, substituting known symbols for these logs, given as \( a \) and \( b \), provides a neat expression: \( \frac{a - b}{b} \). This rewriting not only simplifies computation but also presents the problem in terms familiar and convenient for continued work or specific algebraic tasks.