Problem 13 - Precalculus Mathematics for Calculus

Question

Evaluate the expression. $$\log _{2} 6-\log _{2} 15+\log _{2} 20$$

Step-by-Step Solution

Answer

The expression evaluates to 3.

Step 1: Apply Logarithm Properties

We use the logarithm properties to combine the expression. Recall the property \(\,\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)\,\). Apply this to \(\log_2 6 - \log_2 15\): \[\log_2 6 - \log_2 15 = \log_2 \left(\frac{6}{15}\right)\] This simplifies to \(\log_2 \left(\frac{2}{5}\right)\) since \(\frac{6}{15} = \frac{2}{5}\).

Step 2: Combine Remaining Logarithms

Now we have \(\log_2 \left(\frac{2}{5}\right) + \log_2 20\). Use the property \( \log_b M + \log_b N = \log_b (MN) \) to combine these: \[\log_2 \left(\frac{2}{5}\right) + \log_2 20 = \log_2 \left(\frac{2}{5} \times 20\right)\] Simplify inside the logarithm: \( \frac{2}{5} \times 20 = 8 \).

Step 3: Simplify Expression

We now need to evaluate \(\log_2 8\). Since \(2^3 = 8\), it follows that \(\log_2 8 = 3\). Therefore, the simplified expression is 3.

Key Concepts

Logarithm Properties

Understanding logarithms involves using key properties that simplify expressions. These properties are akin to the rules of arithmetic like addition and subtraction. A fundamental property is the logarithm of a quotients:

• The difference of two logarithms with the same base simplifies to the logarithm of a single quotient: \(\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)\).

Intuitive to understand, this property allows us to condense expressions. Instead of working with multiple terms, you combine them into one. Similarly, when you have a sum, you can use the property:

• The sum of two logarithms with the same base simplifies to the logarithm of a product: \(\log_b M + \log_b N = \log_b (MN)\).

These properties are valuable tools, reducing complex logarithmic expressions to something more manageable.

Simplifying Logarithms

Simplifying logarithms can make evaluating them much easier. In the given exercise, after applying the quotient property of logarithms, we converted two terms into one:\(\log_2 6 - \log_2 15 = \log_2 \left(\frac{2}{5}\right)\).
This simplification results in a single term that is easier to work with. It's like cleaning up clutter—less to focus on and easier to resolve. The smaller terms within the logs often reduce to simpler fractions, which then offer further simplifications.

When continuing, \(\log_2 \left(\frac{2}{5} \times 20\right)\) unfolds into a manageable single term: \(\log_2 8\). This process mirrors condensing numerous tasks into simpler, straightforward actions. Simplification doesn't just tidy up expressions—it clears the path for final evaluations and results.

Logarithmic Operations

Logarithmic operations are the techniques used to manipulate log expressions with addition and subtraction. After simplification using the logarithm properties, the focus shifts to operations.

An example of applying these operations is when we combine multiple logarithmic terms: In the example, \(\log_2 \left(\frac{2}{5}\right) + \log_2 20\), the addition operation allows the combination into a single logarithmic term \(\log_2 8\). This illustrates how operations can transform an expression into something less complex.
Beyond simplifying expressions, these operations help us calculate actual numbers. With \(\log_2 8\), understanding the relationship between powers and bases (where \(2^3 = 8\)) yielded \(\log_2 8 = 3\).

Therefore, logarithmic operations are not just tools for manipulation; they enable us to reach the numeric outcome of a logarithmic expression with clarity.