Problem 26
Question
In Exercises 21–42, evaluate each expression without using a calculator. $$ \log _{6} \frac{1}{6} $$
Step-by-Step Solution
Verified Answer
-1
1Step 1: Express Fraction Using Exponent
Recall that any number \(b\) raised to the power \(-n\) can be written as \(\frac{1}{b^n}\). Therefore, \(\frac{1}{6}\) can be written as \(6^{-1}\)
2Step 2: Apply Logarithmic Property
The base of the logarithm and the base of the exponent are the same. It means \(\log_b{b^n} = n\). Applying this rule, \(\log _{6} 6^{-1} = -1\)
3Step 3: Final Answer
So, \(\log _{6} \frac{1}{6}= -1\)
Key Concepts
ExponentsLogarithmic PropertiesEvaluating Expressions
Exponents
An exponent represents how many times a number, known as the base, is multiplied by itself. For example, in the expression \( 5^3 \), the number 5 is the base, and 3 is the exponent, meaning you multiply 5 by itself three times, resulting in 125. Exponents provide a shorthand way to express repeated multiplication, making it easier to read and understand complex expressions.
When the exponent is a negative number, it signifies taking the reciprocal of the base raised to the corresponding positive exponent. For instance, the expression \( 5^{-2} \) corresponds to \( \frac{1}{5^2} = \frac{1}{25} \).
Exponents are fundamental in the simplification of expressions. They allow us to rewrite fractions and roots in a consistent form, facilitating further operations like logarithms.
When the exponent is a negative number, it signifies taking the reciprocal of the base raised to the corresponding positive exponent. For instance, the expression \( 5^{-2} \) corresponds to \( \frac{1}{5^2} = \frac{1}{25} \).
Exponents are fundamental in the simplification of expressions. They allow us to rewrite fractions and roots in a consistent form, facilitating further operations like logarithms.
Logarithmic Properties
Logarithms are the inverse operations of exponents. They answer the question: "To what power do we need to raise the base to obtain a particular number?" For example, \( \log_2{8} \) asks us to find the power to which 2 must be raised to get 8, and the answer is 3 because \( 2^3 = 8 \).
One key logarithmic property is \( \log_b{b^n} = n \). This property states that if there is a logarithm where the base of the logarithm matches the base of the exponent, the answer is simply the exponent itself. Another important property is \( \log_b{\frac{1}{b}} = -1 \), which follows from the general property for negative exponents that \( b^{-1} = \frac{1}{b} \).
One key logarithmic property is \( \log_b{b^n} = n \). This property states that if there is a logarithm where the base of the logarithm matches the base of the exponent, the answer is simply the exponent itself. Another important property is \( \log_b{\frac{1}{b}} = -1 \), which follows from the general property for negative exponents that \( b^{-1} = \frac{1}{b} \).
- Product Property: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
- Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
- Power Property: \( \log_b(M^p) = p\cdot\log_b(M) \)
Evaluating Expressions
When evaluating mathematical expressions, such as logarithms, it is vital to understand the concepts and properties involved. Evaluating means determining the numerical value of an expression.
In the case of \( \log_6\left(\frac{1}{6}\right) \), we use the knowledge of exponents and logarithmic properties to simplify the expression. Recognizing that \( \frac{1}{6} \) can be expressed as \( 6^{-1} \) allows us to apply the property \( \log_b{b^n} = n \). This simplification leads directly to the result of \( -1 \).
Evaluating expressions, particularly those using logarithms, involves converting the expression into a form that makes it easier to apply known properties. Doing so enables not only solving equations but also gaining a deeper understanding of the relationships between logarithms and exponents.
In the case of \( \log_6\left(\frac{1}{6}\right) \), we use the knowledge of exponents and logarithmic properties to simplify the expression. Recognizing that \( \frac{1}{6} \) can be expressed as \( 6^{-1} \) allows us to apply the property \( \log_b{b^n} = n \). This simplification leads directly to the result of \( -1 \).
Evaluating expressions, particularly those using logarithms, involves converting the expression into a form that makes it easier to apply known properties. Doing so enables not only solving equations but also gaining a deeper understanding of the relationships between logarithms and exponents.
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