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Assume \(\log _{b} x=0.36, \log _{b} y=0.56\) and \(\log _{b} z=0.83 .\) Evaluate the following expressions. $$\log _{b} \frac{x}{y}$$

Short Answer

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Answer: -0.2

Step by step solution

01

Write the given values.

We have the following information given: \(\log _{b} x = 0.36\), \(\log _{b} y = 0.56\), and \(\log _{b} z = 0.83\). We are going to use the values for \(\log _{b} x\) and \(\log _{b} y\) to find \(\log _{b} \frac{x}{y}\).
02

Use the logarithm division rule.

The logarithm division rule states that \(\log _{b} \frac{x}{y} = \log _{b} x - \log _{b} y\). Now, we can plug in the given values for \(\log _{b} x\) and \(\log _{b} y\).
03

Substitute the given values.

Substitute the given values for \(\log _{b} x\) and \(\log _{b} y\). This gives us: \(\log _{b} \frac{x}{y} = 0.36 - 0.56\).
04

Calculate the result.

The final step is to calculate the result of \(\log _{b} \frac{x}{y}\) by executing the subtraction: \(0.36 - 0.56 = -0.2\). The value of \(\log _{b} \frac{x}{y}\) is -0.2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Logarithmic Properties
Logarithms are like the unsung heroes of mathematics, turning multiplication and division into addition and subtraction. To dive into the realm of logarithmic properties, let’s visualize them not just as operations but as translators that convert complicated exponential expressions into a language we can easily understand. Two fundamental properties come into play when you work with logarithms: the multiplication rule and the division rule.

When it comes to multiplying numbers with the same base, the logarithmic property says you can add the logarithms together: \( \log_b(xy) = \log_b(x) + \log_b(y) \). Conversely, when dividing numbers with the same base, as seen in the original exercise, the division rule simplifies it to a subtraction problem: \( \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y) \). Understanding these properties is key to maneuvering through the maze of logarithms with ease.

Once you’ve got these properties down, you’ve essentially learned a new language, enabling you to translate and simplify complex exponential terms into manageable pieces. They allow you to break down the towers of exponents into the bricks of simpler arithmetic operations.
Evaluating Logarithms with Precision
When evaluating logarithms, think of yourself as a detective solving a puzzle. Each log expression you come across has a unique solution based on its base and the number it operates on. Evaluating by hand, as seen in our exercise, involves looking at the given values and applying logarithmic properties accurately.

Evaluating can be straightforward when you have values provided, just like in our case: \( \log_b x = 0.36 \) and \( \log_b y = 0.56 \). But what if you’re faced with expressions without given values or with different bases? This is where your understanding of logarithm properties becomes crucial and why it's important to get comfortable with changing the base of a logarithm or using a calculator effectively for more complicated evaluations.

The process is similar to following a recipe: once you have all your ingredients (values), it's all about mixing them together using the right method—the logarithmic properties.
Demystifying Logarithmic Expressions
Navigating through logarithmic expressions might seem daunting at first glance, but with a bit of practice, it becomes second nature. These expressions are nothing more than a way to represent the power to which a number (the base) must be raised to produce another number. It’s like playing a game of 'Guess the Exponent'.

An expression such as \( \log_b(\frac{x}{y}) \) is a classic example where the division rule of logarithms is used to simplify the expression into a difference of two simpler logarithms. The beauty of logarithmic expressions lies in their diverse applications, from solving exponential equations to modeling real-world phenomena like sound intensity and earthquake magnitudes.

By mastering the art of working with these expressions, you unlock a new toolkit for tackling a vast array of problems in mathematics, physics, and beyond. Remember to approach each expression methodically, peeling away the layers with the help of logarithmic properties until you reach the core value they represent.

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