Evaluate \[$\log_6 \left(\frac{1}{12}\right)\$\].

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Introduction

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In mathematics, logarithms are a fundamental concept that plays a crucial role in various mathematical operations, including algebra, geometry, and calculus. The logarithm of a number is the exponent to which another fixed number, the base, must be raised to produce that number. In this article, we will evaluate the logarithm $\log_6 \left(\frac{1}{12}\right)$, which involves understanding the properties of logarithms and applying them to simplify the expression.

Understanding Logarithms

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A logarithm is the inverse operation of exponentiation. It is a mathematical function that takes a number as input and returns the exponent to which a base number must be raised to produce that number. For example, if we have the equation $2^3 = 8$, then the logarithm of 8 with base 2 is 3, denoted as $\log_2 8 = 3$. This means that 2 raised to the power of 3 equals 8.

Properties of Logarithms

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Logarithms have several properties that are essential in evaluating expressions like $\log_6 \left(\frac{1}{12}\right)$. Some of the key properties of logarithms include:

• Product Rule: $\log_b (xy) = \log_b x + \log_b y$
• Quotient Rule: $\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$
• Power Rule: $\log_b (x^y) = y \log_b x$

Evaluating $\log_6 \left(\frac{1}{12}\right)$

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To evaluate $\log_6 \left(\frac{1}{12}\right)$, we can use the properties of logarithms mentioned above. We can start by expressing $\frac{1}{12}$ as a product of prime factors.

$\frac{1}{12} = \frac{1}{2^2 \cdot 3}$

Step 1: Apply the Quotient Rule

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Using the quotient rule, we can rewrite $\log_6 \left(\frac{1}{12}\right)$ as:

$\log_6 \left(\frac{1}{12}\right) = \log_6 \left(\frac{1}{2^2 \cdot 3}\right)$

Step 2: Apply the Product Rule

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Now, we can apply the product rule to rewrite the expression as:

$\log_6 \left(\frac{1}{2^2 \cdot 3}\right) = \log_6 \left(\frac{1}{2^2}\right) - \log_6 3$

Step 3: Simplify the Expression

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We can simplify the expression further by applying the power rule:

$\log_6 \left(\frac{1}{2^2}\right) - \log_6 3 = -2 \log_6 2 - \log_6 3$

Step 4: Evaluate the Logarithms

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To evaluate the logarithms, we need to find the values of $\log_6 2$ and $\log_6 3$. We can use the change of base formula to rewrite these logarithms in terms of common logarithms:

$\log_6 2 = \frac{\log 2}{\log 6}$

$\log_6 3 = \frac{\log 3}{\log 6}$

Step 5: Substitute the Values

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Substituting the values of $\log_6 2$ and $\log_6 3$ into the expression, we get:

$-2 \log_6 2 - \log_6 3 = -2 \left(\frac{\log 2}{\log 6}\right) - \left(\frac{\log 3}{\log 6}\right)$

Step 6: Simplify the Expression

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Simplifying the expression further, we get:

$-2 \left(\frac{\log 2}{\log 6}\right) - \left(\frac{\log 3}{\log 6}\right) = \frac{-2 \log 2 - \log 3}{\log 6}$

Step 7: Evaluate the Expression

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To evaluate the expression, we need to find the values of $\log 2$, $\log 3$, and $\log 6$. We can use a calculator to find these values:

$\log 2 \approx 0.301$

$\log 3 \approx 0.477$

$\log 6 \approx 0.778$

Step 8: Substitute the Values

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Substituting the values of $\log 2$, $\log 3$, and $\log 6$ into the expression, we get:

$\frac{-2 \log 2 - \log 3}{\log 6} \approx \frac{-2(0.301) - 0.477}{0.778}$

Step 9: Simplify the Expression

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Simplifying the expression further, we get:

$\frac{-2(0.301) - 0.477}{0.778} \approx \frac{-0.602 - 0.477}{0.778}$

Step 10: Evaluate the Expression

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Evaluating the expression, we get:

$\frac{-0.602 - 0.477}{0.778} \approx \frac{-1.079}{0.778}$

Step 11: Simplify the Expression

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Simplifying the expression further, we get:

$\frac{-1.079}{0.778} \approx -1.39$

The final answer is $\boxed{-1.39}$.

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Frequently Asked Questions

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Q: What is a logarithm?

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A: A logarithm is the inverse operation of exponentiation. It is a mathematical function that takes a number as input and returns the exponent to which a base number must be raised to produce that number.

Q: What are the properties of logarithms?

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A: The properties of logarithms include:

• Product Rule: $\log_b (xy) = \log_b x + \log_b y$
• Quotient Rule: $\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$
• Power Rule: $\log_b (x^y) = y \log_b x$

Q: How do I evaluate a logarithm?

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A: To evaluate a logarithm, you can use the properties of logarithms mentioned above. You can start by expressing the number as a product of prime factors and then apply the product and quotient rules to simplify the expression.

Q: What is the change of base formula?

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A: The change of base formula is a formula that allows you to rewrite a logarithm in terms of a common logarithm. It is given by:

$\log_b x = \frac{\log x}{\log b}$

Q: How do I use the change of base formula?

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A: To use the change of base formula, you can substitute the values of $\log x$ and $\log b$ into the formula and simplify the expression.

Q: What is the value of $\log_6 \left(\frac{1}{12}\right)$?

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A: The value of $\log_6 \left(\frac{1}{12}\right)$ is approximately -1.39.

Q: How do I evaluate $\log_6 \left(\frac{1}{12}\right)$?

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A: To evaluate $\log_6 \left(\frac{1}{12}\right)$, you can use the properties of logarithms mentioned above. You can start by expressing $\frac{1}{12}$ as a product of prime factors and then apply the product and quotient rules to simplify the expression.

Q: What are some common logarithms?

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A: Some common logarithms include:

• $\log_2 x$
• $\log_3 x$
• $\log_4 x$
• $\log_5 x$
• $\log_6 x$

Q: How do I use a calculator to evaluate a logarithm?

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A: To use a calculator to evaluate a logarithm, you can enter the expression into the calculator and press the "log" button. The calculator will then display the value of the logarithm.

Q: What are some real-world applications of logarithms?

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A: Logarithms have many real-world applications, including:

• Finance: Logarithms are used to calculate interest rates and investment returns.
• Science: Logarithms are used to calculate the pH of a solution and the concentration of a substance.
• Engineering: Logarithms are used to calculate the gain of an amplifier and the frequency response of a circuit.
• Computer Science: Logarithms are used to calculate the time complexity of an algorithm and the space complexity of a data structure.

Conclusion

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In conclusion, logarithms are a fundamental concept in mathematics that have many real-world applications. By understanding the properties of logarithms and how to evaluate them, you can solve a wide range of problems in finance, science, engineering, and computer science.