Problem 13
Question
Use the Laws of Logarithms to expand the expression. $$ \log _{2}(2 x) $$
Step-by-Step Solution
Verified Answer
The expanded form is \( 1 + \log_{2}(x) \).
1Step 1: Recognize the Logarithmic Property
Notice that the expression \( \log_{2}(2x) \) is in the form of \( \log_{b}(MN) \), where properties of logarithms allow us to expand it into the sum of individual logarithms, \( \log_{b}(M) + \log_{b}(N) \).
2Step 2: Apply the Product Rule of Logarithms
Using the property discovered in Step 1, we apply the product rule: \( \log_{2}(2x) = \log_{2}(2) + \log_{2}(x) \). This breaks the log expression into two separate parts.
3Step 3: Simplify Each Logarithmic Term
Now simplify each part of the expression. We know, \( \log_{2}(2) = 1 \) because 2 to the power of 1 equals 2. Thus, the expression simplifies further to: \( 1 + \log_{2}(x) \). This is the expanded form of the original expression.
Key Concepts
Logarithmic PropertiesProduct Rule of LogarithmsSimplifying Logarithmic Expressions
Logarithmic Properties
Logarithmic properties are fundamental tools that help us manipulate and transform logarithmic expressions. These properties simplify complex expressions, making them easier to solve or expand. The main properties include:
- Product Rule: If you have a logarithm of a product, such as \(\log_{b}(MN)\), you can expand it to \(\log_{b}(M) + \log_{b}(N)\).
- Quotient Rule: If you have a logarithm of a quotient, \(\log_{b}\left(\frac{M}{N}\right)\), it expands to \(\log_{b}(M) - \log_{b}(N)\).
- Power Rule: For the logarithm of a power \(\log_{b}(M^p)\), it becomes \(p\log_{b}(M)\).
Product Rule of Logarithms
The product rule is particularly useful when you're dealing with a logarithm of a product. It states that \(\log_{b}(MN) = \log_{b}(M) + \log_{b}(N)\). This rule allows you to break down complex logarithmic expressions into simpler, more manageable components.
In the given exercise, we had the expression \(\log_{2}(2x)\). Here, the product rule becomes invaluable, allowing us to express it as \(\log_{2}(2) + \log_{2}(x)\). This separable form makes calculations straightforward and reveals insights that might not be immediately visible in the original form.
Using the product rule simplifies not only expansion but also helps when combining different logarithmic expressions into one, making it a versatile tool in solving logarithmic equations.
In the given exercise, we had the expression \(\log_{2}(2x)\). Here, the product rule becomes invaluable, allowing us to express it as \(\log_{2}(2) + \log_{2}(x)\). This separable form makes calculations straightforward and reveals insights that might not be immediately visible in the original form.
Using the product rule simplifies not only expansion but also helps when combining different logarithmic expressions into one, making it a versatile tool in solving logarithmic equations.
Simplifying Logarithmic Expressions
Simplifying logarithmic expressions is a critical skill that lets you reduce complex expressions into more basic forms. Once you've expanded an expression using properties like the product rule, the next step is to simplify each term individually.
For instance, in our example, after applying the product rule, we were left with \(\log_{2}(2) + \log_{2}(x)\). Here, each term is inspected to see if it can be simplified further. We know that \(\log_{2}(2) = 1\) because 2 raised to the power of 1 equals 2. Thus, the expression ultimately simplifies to \(1 + \log_{2}(x)\).
Simplification helps in understanding the underlying values of expressions, solving for unknowns, and making mathematical operations much easier. It's an essential part of mastering logarithms.
For instance, in our example, after applying the product rule, we were left with \(\log_{2}(2) + \log_{2}(x)\). Here, each term is inspected to see if it can be simplified further. We know that \(\log_{2}(2) = 1\) because 2 raised to the power of 1 equals 2. Thus, the expression ultimately simplifies to \(1 + \log_{2}(x)\).
Simplification helps in understanding the underlying values of expressions, solving for unknowns, and making mathematical operations much easier. It's an essential part of mastering logarithms.
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