Problem 1 - Precalculus Mathematics for Calculus

Question

The logarithm of a product of two numbers is the same as the ______ of the logarithms of these numbers. So $$\log _{5}(25 \cdot 125)=$$ _____ + ______ .

Step-by-Step Solution

Answer

The sum of the logarithms, \( \log_5(25) + \log_5(125) \).

Step 1: Identify the logarithmic property

The property used in the logarithm of a product is: \( \log_b(MN) = \log_b(M) + \log_b(N) \). This means the logarithm of a product is the sum of the logarithms of the factors.

Step 2: Apply the property to the problem

Given \( \log_5(25 \cdot 125) \), we can separate it into \( \log_5(25) + \log_5(125) \) based on the product rule identified in Step 1.

Step 3: Fill in the blanks

The problem is asking for the structure \( \log_5(25) + \log_5(125) \). Therefore, the solution is: the sum of the logarithms of these numbers, \( \log_5(25) \) and \( \log_5(125) \).

Key Concepts

Logarithm of a Product

Understanding the concept of "logarithm of a product" is crucial in simplifying complex logarithmic expressions. It states that the logarithm of a product of two numbers is equivalent to the sum of the logarithms of each number. This can be represented mathematically as \( \log_b(MN) = \log_b(M) + \log_b(N) \).

Imagine you have a problem where you need to find the logarithm of the product of two numbers, such as 25 and 125. Instead of calculating \( \log_5(25 \cdot 125) \) directly, you can **break it down** using the product rule:

• Calculate \( \log_5(25) \)
• Calculate \( \log_5(125) \)

This property not only simplifies your calculations but also helps in understanding how logarithmic expressions can be transformed and manipulated effectively.

Sum of Logarithms

The concept behind the "sum of logarithms" stems directly from the logarithm of a product property. When using logarithms in mathematical problems, adding the logarithms is simpler than calculating a single logarithm of a product. This is particularly useful when the numbers are **large or complicated**.

For instance, in our example, converting \( \log_5(25 \cdot 125) \) into a sum of two logarithms, \( \log_5(25) + \log_5(125) \), makes it easier to compute. Each part can be solved independently before summing them. This method saves time and reduces errors:

• Simplifies complex products into smaller calculations
• Breaks down large computations into solvable steps

This technique encourages a deeper understanding of how logarithmic expressions relate to one another and provides a strategic approach to problem-solving.

Precalculus Mathematics

Precalculus mathematics lays the groundwork for more advanced mathematical concepts by introducing primary principles like logarithms. Logarithms are essential in precalculus because they help solve exponential functions, which appear frequently in both natural and applied settings. Understanding their properties, such as the product rule, is fundamental.

As students tackle precalculus, they learn to manipulate and transform expressions by applying these properties effectively. By employing rules like the logarithm of a product and the sum of logarithms, students gain skills necessary for calculus and beyond. Key reasons for mastering these concepts include:

• Building a strong mathematical foundation
• Enabling the simplification of complex expressions
• Preparing for more advanced topics in calculus

Efficient use of logarithmic properties is instrumental in making transitions into calculus smoother and more intuitive for students.