Problem 32 - Algebra A Combined Function

Question

Find the value of each logarithmic expression. $$ \log _{2} \frac{1}{32} $$

Step-by-Step Solution

Answer

The value is -5.

Step 1: Identify the Base and Argument

First, recognize the expression \ \( \log_{2} \frac{1}{32} \) \ means finding the power to which the base 2 must be raised to result in the fraction \( \frac{1}{32} \).

Step 2: Express the Argument as a Power of the Base

Determine how 2 can be used to express \( \frac{1}{32} \). Since \( 32 = 2^5 \), the reciprocal \( \frac{1}{32} = 2^{-5} \).

Step 3: Apply the Definition of Logarithm

According to the properties of logarithms, the expression \( \log_{2}(2^{-5}) \) evaluates to the exponent itself. Therefore, \( \log_{2}(2^{-5}) = -5 \).

Step 4: Conclude the Value

After evaluating the expression, we conclude that \( \log_{2} \frac{1}{32} = -5 \).

Key Concepts

Base and Exponentiation

Logarithms and exponentiation are closely related concepts. To understand a logarithm like \( \log_{2} \frac{1}{32} \), you need to recognize what a base and exponent are. The base in the logarithm is 2 in this case, representing the number you're repeatedly multiplying. The term 'exponent' refers to how many times you multiply the base by itself to achieve a particular value.
In simple terms, exponentiation is defined as:

• \( b^n \) where \( b \) is the base and \( n \) is the exponent.
• For example, \( 2^3 = 2 \times 2 \times 2 = 8 \).

When you see \( \log_{2} \frac{1}{32} \), it asks what power of 2 results in \( \frac{1}{32} \). To solve it, you need to think of \( \frac{1}{32} \) as an exponent expression involving base 2.

Properties of Logarithms

Logarithms are governed by specific properties that make calculating and simplifying them easier. The logarithm \( \log_{b}(x) \) gives us the exponent needed for the base \( b \) to equal \( x \). One crucial property is: when the base of the logarithm and the base of the expression match, you can simplify directly to the exponent:

• If \( b \) is the base of both the logarithm and the result, like \( \log_{2}(2^n) \), it's equal to \( n \).
• This applies to the exercise; \( \log_{2}(2^{-5}) \) simplifies directly to \(-5 \).

Using these properties helps us simplify complex logarithmic expressions without multiplying and dividing repeatedly.
Keep these properties in mind, as they are powerful tools for simplifying many logarithmic problems.

Logarithmic Expressions

Logarithmic expressions can initially seem challenging, but breaking them down step by step simplifies them. Consider any logarithmic expression as a question: what exponent transforms the base into the given value? This makes solving them similar to solving a basic equation. Let's use the exercise to demonstrate:
To find \( \log_{2} \frac{1}{32} \), the key is to rewrite \( \frac{1}{32} \) as a power of 2. This changes the expression into something solvable using properties of exponents. Recognizing that \( 32 = 2^5 \), and since \( \frac{1}{32} \) is the reciprocal, you rewrite it as \( 2^{-5} \).
This transforms your problem into \( \log_{2}(2^{-5}) \), which simplifies to \(-5 \) through simplification rules. In logarithmic terms, this means the power 2 needs to be raised to so that it equals \( \frac{1}{32} \) is \(-5 \). Learning to see logarithmic expressions this way will aid in understanding and solving them more efficiently.