Problem 57 - Algebra for College Students

Question

Given that \(\log _{2} 5=2.3219\) and \(\log _{2} 7=2.8074\), evaluate each expression by using Properties \(10.5-10.7\) \(\log _{2} 175\)

Step-by-Step Solution

Answer

\( \log_{2} 175 = 7.4512 \)

Step 1: Identify the Problem

We need to find the value of \( \log_{2} 175 \). We are given \( \log_{2} 5 = 2.3219 \) and \( \log_{2} 7 = 2.8074 \).

Step 2: Express 175 as a Product of Known Factors

Note that 175 can be expressed as a product: \( 175 = 5 \times 5 \times 7 \). This simplifies to \( 175 = 5^{2} \times 7 \).

Step 3: Apply Logarithm Product Property

Using the product property of logarithms, \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \), we can express \( \log_{2} 175 \) as \( \log_{2} (5^{2} \times 7) = \log_{2}(5^{2}) + \log_{2}(7) \).

Step 4: Apply Logarithm Power Property

Using the power property of logarithms, \( \log_{b}(x^{n}) = n \cdot \log_{b}(x) \), simplify \( \log_{2}(5^{2}) \) to \( 2 \cdot \log_{2}(5) \). The expression becomes \( 2 \cdot \log_{2}(5) + \log_{2}(7) \).

Step 5: Substitute Known Values

Substitute the given values for the logarithms: \( 2 \cdot \log_{2}(5) + \log_{2}(7) = 2 \cdot 2.3219 + 2.8074 \).

Step 6: Calculate the Expression

Calculate \( 2 \cdot 2.3219 = 4.6438 \). Then, add \( 4.6438 + 2.8074 = 7.4512 \).

Key Concepts

Logarithm Properties

Logarithms are mathematical expressions used to simplify operations involving exponents. They have several properties that make calculations, especially involving large numbers, more manageable. Understanding these properties is key to efficiently solving logarithmic expressions.

Some fundamental properties include:

• The product property
• The power property
• The quotient property

The product and power properties are especially useful for expressing complex numbers as simpler components, which we will delve into in more detail.

Product Property of Logarithms

The product property of logarithms states that the logarithm of a product is equal to the sum of the logarithms of the factors. In symbolic terms, it is expressed as: \[\log_b(xy) = \log_b(x) + \log_b(y) \]

This property is incredibly useful when you need to decompose a number into prime factors or smaller components. In the example provided, the number 175 is rewritten as a product of 5 and 7. This allows us to break it down further for more accessible computation. By rewriting 175 as \(5^2 \times 7\), and then using this property, we simplify our calculation to find the value of \(\log_2(175)\).

Power Property of Logarithms

The power property of logarithms involves manipulating exponents within the logarithm, making it simpler. It states that the logarithm of a number raised to an exponent is equal to the exponent multiplied by the logarithm of the base number. Mathematically, this can be expressed as: \[\log_b(x^n) = n \cdot \log_b(x) \]

In practical terms, this property allows you to "bring down" exponents, turning a complex power expression into a simple multiplication problem. For instance, in the provided solution, \(\log_2(5^2)\) becomes \(2 \cdot \log_2(5)\), simplifying the process of calculating the logarithm of products that include powers. This is an excellent example of how mathematical properties can transform complicated problems into straightforward arithmetic.

Evaluating Logarithmic Expressions

Evaluating logarithmic expressions is about applying the right properties to simplify and break down expressions into manageable elements. By using known values, such as given logarithms of numbers, you can easily evaluate more complex expressions.

Returning to the exercise, the process began by rewriting 175 using its prime factorization: \(175 = 5^2 \times 7\). This step is crucial for applying the properties of logarithms effectively. Next, by utilizing the product and power properties, the logarithmic expression was broken down into parts that directly used the provided log values. Finally, by substituting \(\log_2(5) = 2.3219\) and \(\log_2(7) = 2.8074\) into the simplified expression, the calculation became straightforward, resulting in an answer of 7.4512.

This step-by-step use of properties highlights the elegance and power of logarithms in simplifying seemingly daunting arithmetic problems.