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Chapter 4: Problem 41

Use the Laws of Logarithms to combine the $$\log _{2} A+\log _{2} B-2 \log _{2} C$$

Short Answer

Expert verified
Combining the logs gives \( \log_2 \left( \frac{A \cdot B}{C^2} \right) \).

Step by step solution

01

Apply the Product Rule

The product rule of logarithms states that the sum of two logarithms with the same base can be combined into a single logarithm: \[\log_b X + \log_b Y = \log_b (X \cdot Y)\]Here, we can apply the product rule to the first two terms in the expression: \[\log_2 A + \log_2 B = \log_2 (A \cdot B)\]
02

Apply the Power Rule

The power rule of logarithms states that a multiple of a logarithm can be expressed as a power of the argument within a single logarithm: \[c \cdot \log_b X = \log_b (X^c)\]Here, we apply the power rule to the third term in the expression: \[2 \cdot \log_2 C = \log_2 (C^2)\]
03

Apply the Quotient Rule

The quotient rule of logarithms states that the difference of two logarithms with the same base can be combined into a single logarithm: \[\log_b X - \log_b Y = \log_b \left( \frac{X}{Y} \right)\]Now we apply the quotient rule to the modified expression:\[\log_2 (A \cdot B) - \log_2 (C^2) = \log_2 \left( \frac{A \cdot B}{C^2} \right)\]

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Key Concepts

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

Product Rule
The Product Rule is a nifty tool when working with logarithms. It allows you to combine the sum of two logs into a single log by multiplying their arguments. Here's the formula: \( \log_b X + \log_b Y = \log_b (X \cdot Y) \).
Let's break it down:
  • Both logarithms must have the same base.
  • You simply multiply the numbers inside the logs.
  • The result is a single logarithm of the form \( \log_b (X \cdot Y) \).
In our example, we have \( \log_2 A + \log_2 B \). Using the Product Rule, we merge these into \( \log_2 (A \cdot B) \). This simplifies your expression and sets you up nicely for the next step.
Power Rule
The Power Rule is essential when dealing with coefficients in front of logarithms. Essentially, you bring the coefficient inside the logarithm as an exponent of the argument. It looks like this: \( c \cdot \log_b X = \log_b (X^c) \).
Let's dive into it:
  • The base of the log remains the same.
  • Move the coefficient so it becomes an exponent inside the log.
  • Your expression becomes \( \log_b (X^c) \).
In the example, you have \( 2 \cdot \log_2 C \). Applying the Power Rule, it converts to \( \log_2 (C^2) \). This step transforms your logs and makes it easier to deal with them in combination with other rules.
Quotient Rule
The Quotient Rule transforms the difference of two logs into a ratio inside a single log. When you see a subtraction between two logarithms, you can use this rule to simplify: \( \log_b X - \log_b Y = \log_b \left( \frac{X}{Y} \right) \).
Here's how it works:
  • The logarithms should have matching bases.
  • The subtraction means you'll divide the arguments inside the log.
  • You end up with one logarithm, \( \log_b \left( \frac{X}{Y} \right) \).
In our solved expression, you use it on \( \log_2 (A \cdot B) - \log_2 (C^2) \). Applying the Quotient Rule turns this into \( \log_2 \left( \frac{A \cdot B}{C^2} \right) \). This final step simplifies the expression fully, making the logs cleaner and reinforced with these neat rules.

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