Problem 39 - Algebra 2

Question

Evaluate each expression. \(\log _{2} \frac{1}{32}\)

Step-by-Step Solution

Answer

\(-5\)

Step 1: Understanding the Problem

We need to evaluate the logarithmic expression \(\log_{2} \frac{1}{32}\). This expression represents the power to which the base 2 must be raised to yield \(\frac{1}{32}\).

Step 2: Expressing as an Equation

Let's set the given logarithm equal to \(x\), so we have the equation \(2^x = \frac{1}{32}\). This expression means we need to find the value of \(x\) such that when we raise 2 to that power, we get \(\frac{1}{32}\).

Step 3: Rewrite 1/32 as a Power of 2

Recognize that \(32\) is a power of 2. Specifically, \(32 = 2^5\). Thus, we can write \(\frac{1}{32} = \frac{1}{2^5} = 2^{-5}\).

Step 4: Equating the Exponents

Substitute \(\frac{1}{32}\) with \(2^{-5}\) in the equation from Step 2: \(2^x = 2^{-5}\). Since the bases are the same, the exponents must be equal for the equation to hold true, hence \(x = -5\).

Step 5: Conclusion

We have found that the value of \(x\) is \(-5\). Therefore, \(\log_{2} \frac{1}{32} = -5\).

Key Concepts

Logarithmic Expressions

Understanding logarithmic expressions is a key part of mastering logarithms. A logarithmic expression like \( \log_{b}(a) \) essentially asks the question: "To what power must the base \( b \) be raised, in order to get the number \( a \)?" For instance, in the expression \( \log_{2}\left(\frac{1}{32}\right) \), we are asked to find the power that will make the base 2 equal \( \frac{1}{32} \).
Logarithmic expressions are inverses of exponential expressions. They help us solve for exponents when the outcome of the base raised to the power is known.

• Inverse Relation: Allows solving equations involving exponential forms.
• Power Identification: Helps in identifying the required exponent.

Working with logarithmic expressions involves understanding this basic conversion.

• \( b^x = a \rightarrow \log_{b}(a) = x \)

Logarithmic Equations

Logarithmic equations are equations that involve logarithms. They require manipulations similar to those of algebraic equations, but with the properties of logarithms in mind. Consider the equation \( \log_{2}(\frac{1}{32}) = x \).
To solve it, convert the logarithmic form to an exponential form: \( 2^x = \frac{1}{32} \). This step transforms the problem into finding which exponent on 2 results in \( \frac{1}{32} \).

• Converting: Change logarithmic form to exponential form.
• Solving: Find the value of the exponent through equal bases.
• Properties: Use properties of logarithms for simplification.

Solving logarithmic equations often requires patience and practice to recognize patterns.

Exponents

Exponents are a fundamental concept in mathematics, representing repeated multiplication. When you see \( b^x \), it means the base \( b \) is multiplied by itself \( x \) times. In solving \( 2^x = \frac{1}{32} \), recognizing powers and their signs is crucial.
To handle negative exponents, remember that \( b^{-x} = \frac{1}{b^x} \). This notion is evident when solving \( \log_{2}\left(\frac{1}{32}\right) \), as \( 2^{-5} = \frac{1}{32} \).

• Multiplication: Exponents shorten repeated multiplication.
• Negative Powers: Represent division, or reciprocals.
• Zero Power: Any non-zero number to the power of zero is 1 (\( b^0 = 1 \)).

Recognizing patterns in exponents aids in quick computation and understanding.

Base of Logarithms

The base of a logarithm is a crucial part of understanding logarithms completely. It determines the "anchor" with which the powers are counted. In \( \log_{b}(x) \), \( b \) is your base.
In our example, \( \log_{2}\left(\frac{1}{32}\right) \), the base is \( 2 \). This means we use 2 as the number being repeatedly multiplied or divided. The base helps to interpret the logarithm in corresponding exponential terms.

• Common Bases: Base 10 (common log) and base \( e \) (natural log).
• Flexibility: Different bases applicable, but may require conversion.
• Standardization: Choosing a base can standardize calculations.

Knowing the base of the logarithm guides you in both solving and appreciating the behavior of logarithmic functions.