Problem 40 - College Algebra and Calculus: An Applied Approach

Question

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} \frac{3}{5}\)

Step-by-Step Solution

Answer

\(\log _b\frac{3}{5} \approx -0.2625\)

Step 1: Understand the problem

The problem requires calculating \(\log_b (3/5)\) given the values for \(\log _b2\), \(\log _b3\) and \(\log _b5\). This can be achieved by applying the properties of logarithms.

Step 2: Identify the relevant logarithmic property

The property of logarithms that is relevant here is \(\log_b(a/c) = \log_ba - \log_bc\). This property allows the division inside the log to be transformed into a subtraction outside the log, which can further simplified.

Step 3: Apply the property and solve

Applying the property of logarithms, \(\log_b \frac{3}{5} = \log_b3 - \log_b5\), substitute the given values into the equation: \(0.5646 - 0.8271 = -0.2625\). Hence, \(\log _b\frac{3}{5} \approx -0.2625\).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is essential in simplifying complex logarithmic expressions. One of the key properties is that logarithms transform division into subtraction. This is expressed by the formula \( \log_b (\frac{a}{c}) = \log_b a - \log_b c \). This property is extremely useful when you have numerical results of individual parts, just like in our exercise above.

Here’s how it works: When you have a ratio of numbers inside a logarithm, instead of dividing directly, you can find the logs of the numerator and denominator separately and subtract them. It simplifies complex expressions by breaking them down into basic parts that are easier to handle.

Some other useful properties include:

• \( \log_b (a \cdot c) = \log_b a + \log_b c \)
• \( \log_b (a^c) = c \cdot \log_b a \)

Mastering these properties will allow you to manipulate and solve logarithmic problems more efficiently. They're fundamental in simplifying and calculating expressions like those in our exercise.

Logarithmic Approximation

Logarithmic approximation involves estimating logarithmic values using known numbers. While exact values are not always available or calculable, approximations often suffice, especially when precision is not the critical factor. In our exercise, values for \( \log_b 2 \), \( \log_b 3 \), and \( \log_b 5 \) were provided as approximations.

Using the properties of logarithms, namely \( \log_b (\frac{3}{5}) = \log_b 3 - \log_b 5 \), we applied these approximations to solve for \( \log_b (\frac{3}{5}) \). Subtraction of the approximated logs \( 0.5646 - 0.8271 \) yielded \( -0.2625 \). This demonstrates how log approximations are powerful, yet straightforward tools when solving more advanced or abstract expressions in mathematics.

Approximations are widely used in real-world calculations where exact values are complex to determine. Understanding how to leverage this technique efficiently is crucial for quick problem-solving.

Base of Logarithms

The base of a logarithm is the number that is raised to a power to obtain a certain value. In a logarithmic expression \( \log_b a \), \( b \) is the base. The choice of base can affect the complexity and difficulty of calculations. Different bases serve various purposes:

• Common Logarithms: Uses base 10. They are simple and practical, as they correspond to decimal systems.
• Natural Logarithms: Uses base \( e \) (approximately 2.718). These are crucial in calculus and exponential growth models.
• Binary Logarithms: Uses base 2. Often used in computer science due to binary systems.

In our problem, the base \( b \) wasn't specified with a numerical value; however, understanding the base concept is crucial for evaluating how the logarithms combine or reduce. Knowing the base can also facilitate the transition between different log systems, providing flexibility in mathematical operations and problem-solving.