Problem 95 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the exact value of the logarithm without using a calculator. If this is not possible, state the reason.$$\log _{5} 375-\log _{5} 3$$.

Step-by-Step Solution

Answer

The exact value of \( \log _{5} 375 - \log _{5} 3 \) is 3

Step 1: Identify the Logarithmic Difference

Recognize that the expression is a difference between two logarithmic values with the same base. The original expression is as follows: \( \log _{5} 375 - \log _{5} 3 \)

Step 2: Apply the Law of Logarithms

Use the division property of logarithms to express the difference as a single logarithm. According to the law of logarithms, \( \log_b (a) - \log_b (b) = \log_b (a / b) \). Therefore, the expression can be simplified as \( \log _{5} (375 / 3) \)

Step 3: Perform the Division

Divide 375 by 3 to obtain the argument of the logarithm, resulting in \( \log _{5} 125 \)

Step 4: Calculate the Logarithm

Finally, note that 5 raised to the power of 3 equals 125. Thus, \( \log _{5} 125 = 3 \)

Key Concepts

Properties of Logarithms

Understanding the fundamental properties of logarithms is crucial for solving logarithmic equations and manipulating expressions. A logarithm, written as \( \text{log}_b(a) \), essentially asks the question 'To what power must the base \( b \) be raised, to produce the number \( a \)?'

There are three main properties that can greatly simplify logarithmic calculations:

• Product Property: \( \text{log}_b(mn) = \text{log}_b(m) + \text{log}_b(n) \), which means the logarithm of a product is equal to the sum of the logarithms of the factors.
• Quotient Property: \( \text{log}_b(\frac{m}{n}) = \text{log}_b(m) - \text{log}_b(n) \), indicating that the logarithm of a quotient is the difference between the logarithms of the numerator and the denominator.
• Power Property: \( \text{log}_b(m^n) = n \text{log}_b(m) \), which tells us that the logarithm of a power is the exponent times the logarithm of the base.

Applying these properties to logarithmic expressions can transform seemingly complicated equations into simpler components that are easier to solve. As students become more familiar with these properties, logarithm calculations will become a matter of recognizing patterns and applying the appropriate rules.

Logarithmic Difference

The concept of logarithmic difference harnesses the quotient property of logarithms and provides a method to subtract two logs with the same base. In the context of the given exercise, you're faced with \( \text{log}_{5}(375) - \text{log}_{5}(3) \). This is not just a random subtraction of numbers; it follows a specific logarithmic rule that simplifies the process.

Why do we subtract logarithms? Often, subtraction surfaces naturally when we're dealing with ratios or proportions in variables, resulting in a logarithmic difference. Instead of working with two separate logarithmic terms, we can combine them into a single term that represents the log of a division, using the above-mentioned quotient property. This simplification can lead to an easier path to finding the values of logarithms, especially when the resulting value, like in the exercise, is a known power of the base.

Division Property of Logarithms

The division property of logarithms is a powerful tool, particularly when it comes to simplification of logarithmic expressions. To reiterate, the quotient property states: \( \text{log}_b(\frac{m}{n}) = \text{log}_b(m) - \text{log}_b(n) \).

This property is essentially used to rewrite the logarithm of a division as the difference of two logarithms. Why is this helpful? It allows you to break down complex expressions into smaller, more manageable parts.

In the context of the exercise \( \text{log}_{5}(375) - \text{log}_{5}(3) \), it enables us to condense the expression to \( \text{log}_{5}(125) \), a single logarithmic term. Recognizing that 125 is a power of 5, we can directly identify the exact value of the logarithm without using a calculator, leading to the solution \( \text{log}_{5}(125) = 3 \). This not only streamlines calculations but also reinforces the relationship between exponentiation and logarithms as inverse operations.