Problem 51 - Intermediate Algebra

Question

For Problems \(51-59\), you are given \(\log _{2} 5=2.3219\) and \(\log _{2} 7=2.8074\). Evaluate each expression using Properties 11.5-11.7. $$ \log _{2} 35 $$

Step-by-Step Solution

Answer

\( \log_{2} 35 = 5.1293 \).

Step 1: Identify the Expression

The given expression is \( \log_{2} 35 \). We need to evaluate this using the properties of logarithms.

Step 2: Use the Product Rule

According to the product rule of logarithms, \( \log_{b}(MN) = \log_{b}M + \log_{b}N \). This allows us to express \( \log_{2} 35 \) as \( \log_{2} (5 \times 7) \). Therefore, \( \log_{2} 35 = \log_{2} 5 + \log_{2} 7 \).

Step 3: Substitute Known Values

Substitute the known values into the equation: \( \log_{2} 5 = 2.3219 \) and \( \log_{2} 7 = 2.8074 \).

Step 4: Perform the Addition

Calculate the sum: \( 2.3219 + 2.8074 = 5.1293 \).

Step 5: Result

Therefore, \( \log_{2} 35 \) evaluates to \( 5.1293 \).

Key Concepts

Product Rule of Logarithms

One of the most powerful tools when working with logarithms is the product rule. This rule simplifies multiplication inside a logarithm by transforming it into an addition statement. The rule states:

• \( \log_{b}(MN) = \log_{b}M + \log_{b}N \)

Essentially, if you have a product within the log you can break it into a sum of logs. For instance, evaluating \( \log_{2} 35 \) might seem complex at first glance due to the multiplication inside the logarithm. However, using the product rule, we can think of 35 as \( 5 \times 7 \) and break it down as \( \log_{2} 5 + \log_{2} 7 \).
Breaking down big numbers into smaller factors and applying the product rule makes it easier to evaluate and work with logarithmic expressions.

Logarithm Properties

Logarithms have several unique and helpful properties that make dealing with complex expressions much more manageable. In our previous example, we utilized the product rule. Here are a few more essential properties:

• Power Rule: \( \log_{b}(M^{n}) = n \cdot \log_{b}M \) - allows you to bring the exponent before the log.
• Quotient Rule: \( \log_{b}\left( \frac{M}{N} \right) = \log_{b}M - \log_{b}N \) - helps break down logarithms of fractions.
• Change of Base Formula: \( \log_{b} M = \frac{\log_{k} M}{\log_{k} b} \) - useful when you want to convert logs to different bases.

These properties give you flexibility and can make seemingly daunting calculations straightforward. Delving into these properties, as demonstrated in the exercise, builds a strong foundation for solving logarithmic expressions efficiently.

Evaluating Logarithmic Expressions

Evaluating logarithmic expressions involves applying known properties and values to simplify and find numeric results. In practical terms, it means using rules like the product rule and substituting known values into your expression.
For our specific problem, evaluating \( \log_{2} 35 \) required breaking it down into its factors using the product rule. Once broken down, you substitute in the given values: \( \log_{2} 5 = 2.3219 \) and \( \log_{2} 7 = 2.8074 \).
The next step is straightforward arithmetic: adding these values to achieve the result. This evaluation process shows the effectiveness of the logarithm properties in simplifying and computing results quickly and accurately. Whether you’re dealing with base 2 logs like in this example or using other bases, the underlying principles remain consistent.