Problem 35 - Intermediate Algebra

Question

For Problems \(21-40\), evaluate each expression. $$ \log _{2}\left(\frac{1}{32}\right) $$

Step-by-Step Solution

Answer

The answer is \( -5 \).

Step 1: Understand the Logarithm Definition

Recall the definition of the logarithm: If \( \log_b(a) = c \), then \( b^c = a \). We need to find the exponent \( c \) such that \( 2^c = \frac{1}{32} \).

Step 2: Convert to a Fractional Exponent

Recognize that \( \frac{1}{32} \) is the same as \( 32^{-1} \). We know \( 32 = 2^5 \), so \( \frac{1}{32} = (2^5)^{-1} = 2^{-5} \).

Step 3: Set the Equation and Solve for the Logarithm

We now have the equation \( 2^c = 2^{-5} \) where \( c \) is the unknown we are solving for. Since the bases are the same, set the exponents equal: \( c = -5 \).

Step 4: Conclusion

The value of \( \log _{2}\left(\frac{1}{32}\right) \) is \(-5\).

Key Concepts

Logarithm Properties

Logarithms are mathematical expressions that provide a way to solve for unknown exponents. They are the inverse operations of exponentiation. Let's explore some key properties of logarithms to deepen our understanding. These properties are useful when simplifying and solving logarithmic expressions.

• Product Property: The logarithm of a product is the sum of the logarithms, i.e., \( \log_b(MN) = \log_b(M) + \log_b(N) \).
• Quotient Property: The logarithm of a quotient is the difference of the logarithms, \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \).
• Power Property: The logarithm of a power allows us to bring the exponent in front, \( \log_b(M^k) = k \cdot \log_b(M) \).

These properties simplify complex logarithmic expressions and enable us to solve logarithmic equations efficiently. Knowing these can help you break down logarithmic problems into simpler forms that are easier to solve.

Fractional Exponents

Fractional exponents, also known as rational exponents, represent another way to express roots and powers in mathematics. Understanding fractional exponents is crucial to solving expressions like the logarithmic problem above. Let's break down what they mean and how they work.

A fractional exponent \( a^{m/n} \) denotes both an exponent and a root:

• The numerator \( m \) signifies the power.
• The denominator \( n \) signifies the root, such that \( a^{m/n} = \sqrt[n]{a^m} \).

For instance, an expression like \( 32^{-1} \), as shown in the problem, can be rewritten using fractional exponents. Here, \( 32 = 2^5 \), so \( 32^{-1} \) equates to \((2^5)^{-1} = 2^{-5} \). Fractional exponents are a versatile tool for simplifying expressions and solving equations that involve powers and roots. Recognizing and manipulating these can greatly aid in finding solutions to problems involving exponents.

Base 2 Logarithms

When working with logarithms, the base indicates which number's powers we are considering. For our problem, the logarithm is in base 2, which plays a significant role in simplifying the expression. Here's why base 2 logarithms are important and how they can be interpreted and used effectively.

Base 2 logarithms (\( \log_2 \)) are commonly encountered in fields like computer science and engineering because binary (base 2) systems are fundamental to computational processes. Logarithms predict how many times you multiply 2 to reach a certain value.

In our exercise, evaluating \( \log_2\left(\frac{1}{32}\right) \) involves determining the power of 2 that results in \( \frac{1}{32} \). We expressed \( \frac{1}{32} \) as \( 2^{-5} \) because \( 32 = 2^5 \), meaning \( \log_2\left(\frac{1}{32}\right) = -5 \).

Understanding base 2 logarithms helps solve exponentiation problems efficiently, especially when dealing with halves, doubles, and powers related to 2. This makes interpreting such logarithms an essential skill for anyone working with exponential concepts.