Problem 14 - Intermediate Algebra

Question

Match each expression with an equivalent expression from the list on the right. a. \(\frac{\log 11}{\log 3}\) b. \(11 \log _{3} 10\) c. \(\log _{3} 5+\log _{3} 2\) d. \(\log _{3} 10-\log _{3} 11\) $$ \log _{3} 11 $$

Step-by-Step Solution

Answer

Expression (a) matches with \( \log_{3} 11 \). None of the other expressions in the list match directly with \( \log_{3} 11 \).

Step 1: Identify Equivalent Logarithmic Expression (a)

You are given the expression \( \frac{\log 11}{\log 3} \). This can be rewritten using the change of base formula as \( \log_{3} 11 \), because \( \log_b a = \frac{\log_c a}{\log_c b} \). So, a matches with \( \log_{3} 11 \).

Step 2: Analyze Expression (b)

The expression \( 11 \log_{3} 10 \) can be assessed for equality with any in the list. A logarithm property states \( \log_b (a^c) = c \log_b a \). There is no direct match with \( \log_3 11 \) unless transformed further by context, implying no equivalent match directly on the list.

Step 3: Simplify Expression (c)

The expression \( \log_3 5 + \log_3 2 \) uses the property of logarithms that states \( \log_b a + \log_b c = \log_b (ac) \). Therefore, this equals \( \log_3 (5 \times 2) = \log_3 10 \) which does not directly match \( \log_3 11 \). Hence, there's no equal match for this one from the given list.

Step 4: Simplify Expression (d)

The expression \( \log_3 10 - \log_3 11 \) utilizes the logarithmic property \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \). Therefore, it simplifies to \( \log_3 \left( \frac{10}{11} \right) \). No direct equivalent with \( \log_3 11 \) is available for the given choices.

Key Concepts

Change of Base Formula

The change of base formula is an essential tool when dealing with logarithms, especially when you want to convert between different bases.
It's just like changing a foreign currency into your local currency to understand its value better.
In logarithms, the change of base formula states that:

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

Here, you can convert a logarithm of any base \(b\) to another base \(c\). The numerators and denominators become the logarithms in your new base \(c\).
For instance, to convert \(\log_{3} 11\) into a more common base, like base 10, you can rewrite it as \(\frac{\log 11}{\log 3}\). This is especially useful when using calculators that only handle logarithms in base 10 or base \(e\) (natural logarithms).
This formula not only helps in calculations but also in simplifying problems where expressions need to be matched, as seen in the exercise.

Logarithmic Properties

Understanding logarithmic properties can significantly ease the process of simplifying and manipulating logarithmic expressions. These properties are rules governing how logs behave and relate to each other.
One fundamental property to consider is the logarithm of a product:

• \( \log_b(a \cdot c) = \log_b a + \log_b c \)

This property allows you to split a multiplication inside the log into addition, breaking down the expression for easier understanding.
Another important property is the logarithm of a quotient:

• \( \log_b\left(\frac{a}{c}\right) = \log_b a - \log_b c \)

This property helps convert divisions within the log into a subtraction between two logs. Both properties are vital when you're tasked with simplifying log expressions or verifying their equivalence.
For instance, in our exercise, the expression \(\log_3 5 + \log_3 2\) becomes \(\log_3 10\) using the product property. Similarly, \(\log_3 10 - \log_3 11\) simplifies through the quotient property to \(\log_3\left(\frac{10}{11}\right)\). Being familiar with these straightforward rules makes working with logarithms much more manageable.

Algebraic Matching

Matching logarithmic expressions involves the mathematical art of recognizing equivalences and transformations.
It's a bit like solving a puzzle: looking for pieces that fit based on given rules.
To effectively perform algebraic matching, one must utilize both the change of base formula and logarithmic properties.
Let's consider what happens with our example: \(\frac{\log 11}{\log 3}\) matches with \(\log_{3} 11\) using the change of base formula. This demonstrates how seemingly different expressions in different bases can represent the same value.
Next, consider how using properties of logarithms like the product and quotient rules helps match or simplify expressions, even when they don't have a direct pair.
For instance:

• \(\log_3 5 + \log_3 2\) translates to \(\log_3 10\), though unmatched here, it's important to identify such transformations for complex problems.

Being able to move smoothly between these forms is a powerful skill. It aids in clearer problem-solving and better understanding in algebra and beyond.