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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}(2 b)^{-2}\)

Short Answer

Expert verified
The approximation of \( \log _{b}(2 b)^{-2} \) is -2.7124.

Step by step solution

01

Simplify the exponent in the logarithm

The first step is to use the power rule of logarithms, which states that \( \log_{b}A^n = n*\log_{b}A \). Applying this to \( \log_{b}(2b)^{-2} \), it simplifies to: \( -2*\log_{b}(2b) \).
02

Break down the logarithm

The second step is to use another property of logarithms, the product rule, which states that \( \log_{b}AB = \log_{b}A + \log_{b}B \). Hence, \( \log_{b}(2b) \) simplifies to: \( \log_{b}2 + \log_{b}b \).
03

Replace known logarithm values

Replace \( \log_{b}2 \) with 0.3562. Also, note that \( \log_{b}b \) is equal to 1, since any base logarithm of that same base equals 1. The expression now becomes -2*(0.3562 + 1).
04

Compute the approximate value

Calculate -2*(0.3562 + 1) which yields -2.7124. Thus, the approximation of \( \log _{b}(2 b)^{-2} \) is -2.7124.

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Key Concepts

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

Understanding Properties of Logarithms
Logarithms have a set of unique properties that simplify complex calculations and make problem-solving easier. These properties allow us to manipulate logarithmic expressions in ways that can reduce them to more manageable forms. Here are some of the most helpful properties:
  • Product Rule: This states that the logarithm of a product is equal to the sum of the logarithms of its factors. Mathematically represented as \( \log_{b}(AB) = \log_{b}A + \log_{b}B \).
  • Power Rule: This allows us to bring down the exponent in logarithmic expressions. Represented as \( \log_{b}A^n = n \times \log_{b}A \).
  • Base Rule: Any logarithm of a base to itself equals 1, meaning \( \log_{b}b = 1 \).

These properties are essential tools, particularly when analyzing or approximating logarithms in educational math problems.
Applying the Power Rule
In mathematics, the power rule is a fundamental concept used in manipulating logarithmic expressions. The power rule states that if you have a log of something raised to a power, you can bring that power in front as a multiplier. For example, given an expression \( \log_{b}(A^n) \), this simplifies to \( n \times \log_{b}(A) \).
In the original exercise, the problem \( \log_{b}(2b)^{-2} \) uses this rule. Since \(-2\) is an exponent, according to the power rule, it can be moved in front of the logarithm. Therefore, it translates to \(-2 \times \log_{b}(2b)\), making the expression simpler to manage. Understanding and applying this rule is crucial when dealing with powers in logarithmic expressions.
Understanding the Product Rule
The product rule is another valuable property of logarithms. This rule is about breaking down the logarithm of a product into a sum of logarithms. When you have a logarithm like \( \log_{b}(AB) \), the product rule lets you split it into \( \log_{b}A + \log_{b}B \).
In the context of our exercise, we were given \( \log_{b}(2b) \). By applying the product rule, it becomes \( \log_{b}2 + \log_{b}b \). Given that \( \log_{b}b = 1 \), we can quickly evaluate this expression.
The product rule helps break down complex logarithmic expressions, making them simpler and more accessible.
Solving an Educational Math Problem
Tackling logarithmic problems like the one in the original exercise can be made simpler with a step-by-step approach. In this exercise, we need to approximate the value of \( \log_{b}(2b)^{-2} \) using given values for \( \log_{b}2 \), \( \log_{b}3 \), and \( \log_{b}5 \).
By first applying the power rule, we transform the original expression to \(-2 \times \log_{b}(2b)\). Next, using the product rule, \( \log_{b}(2b) \) becomes \( \log_{b}2 + \log_{b}b \). Since \( \log_{b}2 \approx 0.3562 \) and \( \log_{b}b = 1 \), we substitute these values, resulting in \(-2 \times (0.3562 + 1)\). Upon simplifying, the calculation gives approximately \(-2.7124\).
This methodical approach highlights the importance of understanding and applying logarithm properties to solve and approximate logarithmic problems efficiently.

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