Problem 12 - Algebra 2

Question

Use \(\log _{5} 2 \approx 0.4307\) and \(\log _{5} 3 \approx 0.6826\) to approximate the value of each expression. \(\log _{5} 50\)

Step-by-Step Solution

Answer

\( \log_5 50 \approx 2.4307 \)

Step 1: Express 50 in terms of 2 and 3

Recognize that 50 can be expressed as a product of 2, 3, and a power of 5. Start by factorizing 50 into its prime factors: \ \[ 50 = 2 imes 5^2 \] \ This expression uses 2 and the base 5.

Step 2: Apply Logarithm Product Rule

Use the logarithm product rule, which states \( \log_b (mn) = \log_b m + \log_b n \). Apply this to the expression: \ \[ \log_5 (2 \times 5^2) = \log_5 2 + \log_5 5^2 \] \ We need to break it down into its components.

Step 3: Simplify Using Known Logarithms

Since \( \log_5 5^2 = 2 \log_5 5 \) and knowing that \( \log_5 5 = 1 \), we can further simplify: \ \[ \log_5 5^2 = 2 \times 1 = 2 \] \ Then, the expression becomes: \ \[ \log_5 2 + 2 \]

Step 4: Substitute Known Values and Calculate

Substitute the known approximate value \( \log_5 2 \approx 0.4307 \) into the equation: \ \[ 0.4307 + 2 = 2.4307 \] \ Thus, \( \log_5 50 \approx 2.4307 \).

Key Concepts

Logarithm Product Rule

Logarithms are quite fascinating as they allow us to simplify the multiplication of numbers through addition. This can be efficiently done using the **logarithm product rule**. The rule states that for any positive numbers \(m\) and \(n\), and any base \(b\), the logarithm of a product is equal to the sum of the logarithms:

• \( \log_b(mn) = \log_b m + \log_b n \)

This rule is incredibly useful when we need to break down a complex expression into simpler parts. In the context of the original problem, where we wish to find \( \log_5 50 \), representing 50 as a product helps. Once 50 is expressed in terms of prime factors (\(50 = 2 \times 5^2\)), we apply the product rule:

• \( \log_5 (2 \times 5^2) = \log_5 2 + \log_5 5^2 \)

This enables us to deal with components we might already know, making calculations simpler. For instance, if we know \( \log_5 2 \), we can easily plug it into the expression.

Approximation of Logarithms

Approximating logarithms means finding a close numerical estimate of a logarithmic expression. It is a handy technique, especially when exact values are not readily available or when simplifying calculations. In our exercise, we're given approximations like \( \log_5 2 \approx 0.4307 \). Approximations are useful because:

• They provide manageable numbers for calculations.
• They can be used in various mathematical and practical applications where precision is flexible.

Approximations are derived from known values or through methodologies like interpolation. In handling our problem, once 50 is expressed using known primes, we substitute the known approximate value \( \log_5 2 \approx 0.4307 \) into the expression. This is done after dealing with \( \log_5 5^2\) which simplifies to 2, since \( \log_5 5 = 1 \).
This step highlights how approximations blend theoretical understanding with practical computations, yielding results efficiently without diving into complex calculations.

Prime Factorization

Prime factorization breaks down a number into a set of basic building blocks, which are prime numbers. Every integer has a unique prime factorization. This is essential in multiple areas of mathematics, including the simplification of logarithmic expressions.
To solve \( \log_5 50 \), first we need to recognize that 50 can be rewritten as:

• \( 50 = 2 \times 25 = 2 \times 5^2 \)

This prime factorization is key as it helps reduce the expression into parts that are manageable using the logarithm product rule. Recognizing the power of primes like 5 here allows us to simplify \( \log_5 5^2 \) directly. Each prime factor contributes to a term in the logarithmic expansion, enabling us to compute logarithms through more straightforward arithmetic operations.
This concept of breaking a number into its prime constituents doesn't just facilitate complex calculations but also lays a foundational understanding necessary for exploring number theory and computational mathematics.