Problem 12 - 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. \(\log _{3} 10-\log _{3} 11\) $$ \log _{3} \frac{10}{11} $$

Step-by-Step Solution

Answer

Expression (d) matches \(\log_3 \frac{10}{11}\).

Step 1: Understanding Definitions

Recall the properties of logarithms: \(\log_a b = \frac{\log_c b}{\log_c a}\) and \(\log_a (b\cdot c) = \log_a b + \log_a c\). Also, notice that \(\log_a\left(\frac{b}{c}\right) = \log_a b - \log_a c\).

Step 2: Matching Expression with Equivalent

Look at expression (d) \(\log _{3} 10-\log _{3} 11\). This is an application of the subtraction (division) property of logarithms: \(\log_3\left(\frac{10}{11}\right)\), which matches with the given expression \(\log_3 \frac{10}{11}\).

Step 3: Verify Other Expressions

For completeness, check other expressions: - (a) \(\frac{\log 11}{\log 3}\) can be rewritten using the change of base formula as \(\log_3 11\). - (b) \(11 \log_3 10\) is unrelated to \(\log_3 \frac{10}{11}\). - (c) \(\log_3 5 + \log_3 2\) can be rewritten as \(\log_3 (5\times 2) = \log_3 10\). None of these directly match \(\log_3 \frac{10}{11}\) except (d).

Step 4: Conclusion

(d) \(\log _{3} 10-\log _{3} 11\) is equivalent to \(\log _{3} \frac{10}{11}\). Hence, it's the matched expression.

Key Concepts

Properties of Logarithms

Logarithms have special properties that make calculations much easier. These properties can transform complex expressions into simpler forms. Here are the main ones:

• Product Property: When you multiply two numbers, the logarithm of the result is the addition of their logarithms. Mathematically, it is expressed as: \[\log_a(b \cdot c) = \log_a(b) + \log_a(c)\]
• Quotient Property: When dividing two numbers, the logarithm of the result is the difference between the two logarithms:\[\log_a\left(\frac{b}{c}\right) = \log_a(b) - \log_a(c)\]
• Power Property: If a number is raised to an exponent, the logarithm of the result is the exponent times the logarithm of the base number:\[\log_a(b^n) = n \cdot \log_a(b) \]

These properties are like tools in your math toolkit, helping to rewrite or simplify logarithmic expressions easily. If you grasp these fundamentals, manipulating logarithms becomes much more straightforward.

Change of Base Formula

Calculating logarithms with a non-standard base can be tricky, but the change of base formula makes it easy. This formula allows you to convert a logarithm to any base you prefer, often making calculations simpler. Here's how it goes:\[\log_a b = \frac{\log_c b}{\log_c a}\]This expression implies that you can compute the logarithm of a number with any base, using a more convenient base, like 10 or e (the natural logarithm base). Why is this useful?

• It simplifies calculations when you only have a limited set of logarithm tables or calculators, usually equipped with base 10 or \( e \).
• It provides flexibility in solving complex logarithmic expressions, by converting them to a form that aligns with your needs or available resources.

The formula takes advantage of the availability of certain computational tools, allowing for more effective problem solving. Understanding and leveraging the change of base formula makes dealing with non-standard bases much easier.

Division Property of Logarithms

The division property of logarithms allows you to express a logarithm as the subtraction of two others. This property is really helpful when working with division in logarithmic expressions. It's expressed as:\[\log_a\left(\frac{b}{c}\right) = \log_a(b) - \log_a(c)\]This property simplifies the work with logarithms by allowing you to convert a division inside a logarithm into a subtraction problem.Consider the example from our exercise, \( \log_3 10 - \log_3 11 \). Using the division property, this can be rewritten as \( \log_3\left(\frac{10}{11}\right) \), showing clearly how subtraction inside logarithms handles division problems.

• It helps in simplifying equations involving division, making them easier to solve.
• It's particularly useful when matching complex expressions by directly breaking them down into simpler components.

The division property is a valuable ally, enabling you to handle tricky divisions within logarithmic functions by simplifying them into basic operations.