Problem 62 - Algebra for College Students

Question

Given that \(\log _{8} 5=0.7740\) and \(\log _{8} 11=1.1531\), evaluate each expression using Properties \(10.5-10.7\) \(\log _{8} 25\)

Step-by-Step Solution

Answer

\( \log_8 25 = 1.5480 \)

Step 1: Identify the Property to Use

To solve for \( \log_8 25 \), we need a property of logarithms that allows us to relate known logarithms to the expression we want to evaluate. In this case, we use the Product Property of Logarithms which states that \( \log_b(MN) = \log_b M + \log_b N \). So, we can express \( 25 \) as \( 5 \times 5 \) and apply this property.

Step 2: Apply the Product Property

Using the Product Property, we can write \( \log_8 25 \) as \( \log_8 (5 \times 5) \), which equals \( \log_8 5 + \log_8 5 \).

Step 3: Substitute Known Values

We know \( \log_8 5 = 0.7740 \). Now substitute this value into the expression from Step 2: \( \log_8 25 = 0.7740 + 0.7740 \).

Step 4: Perform the Calculation

Add the values obtained in Step 3: \( 0.7740 + 0.7740 = 1.5480 \). This is the value of \( \log_8 25 \).

Key Concepts

Product Property of Logarithms

In the world of logarithms, the product property is one of the key tools that we use to simplify expressions or solve problems. The product property states:

• If you have two numbers, say \( M \) and \( N \), and you want to find the logarithm of their product with the same base \( b \), you can add the logarithms of the individual numbers: \( \log_b(M \cdot N) = \log_b M + \log_b N \).

This property is incredibly useful because it allows us to break down complex expressions into simpler parts by leveraging the relationships between numbers.

For example, if you know the logarithm values of smaller numbers, you can deduce the logarithm for larger numbers by breaking them down into products of the smaller numbers. In our exercise, we used it by expressing \( 25 \) as \( 5 \times 5 \), allowing us to evaluate \( \log_8 25 \) using the known value of \( \log_8 5 \).

Logarithmic Expressions

Logarithmic expressions represent the power to which a fixed number, the base, must be raised to produce a given number. They are written as \( \log_b N \), where \( b \) is the base and \( N \) is the number.

Key aspects of working with logarithmic expressions include:

• Understanding the base, which dictates the number that is repeatedly multiplied.
• Knowing how to convert between logarithmic and exponential forms.
• Utilizing properties of logarithms, such as the product, quotient, and power laws to simplify expressions.

When dealing with logarithmic expressions, it's often necessary to manipulate them to match known values or to simplify calculations. In our exercise, by expressing \( 25 \) as \( 5 \times 5 \) and applying the product property, we were able to use known values to easily calculate \( \log_8 25 \).

Algebraic Problem-Solving

Algebraic problem-solving involves using properties and rules of algebra to simplify and solve mathematical problems. Logarithms often feature in algebraic problems because of their ability to transform multiplicative relationships into additive ones.

In algebraic problem-solving:

• Identify patterns or properties that simplify complex expressions.
• Transform expressions using known mathematical properties, like the product property for logarithms.
• Substitute known values to find unknowns or simplify calculations.

For example, in our provided exercise, the problem was to evaluate \( \log_8 25 \) using given values. Using the product property, we broke down \( 25 \) into two factors, \( 5 \times 5 \), enabling us to substitute the known value for \( \log_8 5 \) into our expression. Finally, through simple arithmetic, we achieved an easy solution for the problem, highlighting the effectiveness of algebraic problem-solving.