Problem 14 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The equivalent expression is (a) \(\frac{\log 11}{\log 3}\).

Step 1: Understand logarithm conversion

The expression \(\log_{3} 11\) means the logarithm of 11 with base 3. In mathematics, there is a property of logarithms known as the change of base formula. According to the change of base formula, \(\log_{b} a = \frac{\log_{c} a}{\log_{c} b}\) for any positive number \(c\), where \(c eq 1\). This means that a logarithm with any base can be converted using natural logarithms or common logarithms.

Step 2: Apply the change of base formula

Apply the change of base formula to the expression \(\log_{3} 11\). Using the common logarithm (base 10), it can be rewritten as follows: \(\log_{3} 11 = \frac{\log 11}{\log 3}\). This allows us to match \(\log_{3} 11\) with option (a) \(\frac{\log 11}{\log 3}\).

Key Concepts

Change of Base Formula

In the realm of logarithms, the change of base formula is a powerful tool. It allows the conversion of logarithms from one base to another. This is incredibly useful, especially when your calculator only supports common (base 10) or natural (base e) logarithms. The formula can be expressed as \(\log_{b} a = \frac{\log_{c} a}{\log_{c} b}\), where \(b\) is the original base, \(a\) is the number we're taking the logarithm of, and \(c\) is the new base.

• The key to using this formula is picking a new base, \(c\), that simplifies the problem or makes computation possible on a standard calculator.
• Most commonly, we choose \(c\) to be either 10 or \(e\), as these are the bases for the common logarithm (logarithm base 10) and natural logarithm (logarithm base \(e\)), respectively.

Let’s see it in action: applying the change of base formula to \(\log_{3} 11\) with base 10 gives
\(\log_{3} 11 = \frac{\log 11}{\log 3}\). This expression is now in a form that can be easily evaluated using a calculator that supports common logarithms.

Logarithm Conversion

Logarithm conversion is the process of transforming a logarithm from one base to a potentially more manageable base. This often involves the use of the change of base formula for easing computation. Here are some important points to remember about logarithm conversion:

• The essence of conversion is to make problems approachable when dealing with complex bases.
• For practical use, most calculations in algebra, trigonometry, or calculus are done using base 10 or base \(e\), because these are supported by most calculators.
• Conversion is not just a formality - it can drastically simplify the arithmetic involved when calculating logarithmic expressions.

For instance, converting \(\log_{3} 11\) using the common logarithm, the expression becomes \(\frac{\log 11}{\log 3}\), which is much simpler to compute. This equivalent expression has the same value but is more straightforward to calculate.

Equivalent Expressions

Equivalent expressions in mathematics refer to different forms of an expression that yield the same value. In the context of logarithms, these equivalent forms often arise after performing operations such as applying the change of base formula.

Let's discuss this in terms of logarithms: When you convert \(\log_{3} 11\) into \(\frac{\log 11}{\log 3}\), you produce an equivalent expression. Despite looking different, it retains the same value.

• Such conversions and transformations are crucial because they can simplify problems or make them solvable with available tools.
• Recognizing equivalent expressions can be highly beneficial for solving equations, evaluating complex expressions, and simplifying calculations.
• Practicing the recognition and conversion into equivalent expressions will equip you with strategies to handle various algebraic challenges.

By understanding and applying these principles, you can ensure that you are working with expressions in the most efficient and effective way possible.