Problem 16 - College Algebra with Modeling and Visualization

Question

Evaluate the expression by hand. $$ \left(2^{-2}\right)^{-3 / 2} $$

Step-by-Step Solution

Answer

The evaluated expression is 8.

Step 1: Simplify the Inner Exponent

Start by simplifying the inner expression, which is \( 2^{-2} \). Recall the rule of exponents \( a^{-n} = \frac{1}{a^n} \). Thus, \( 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \).

Step 2: Apply the Outer Exponent

Next, we apply the outer exponent \( -\frac{3}{2} \) to the simplified form of \( 2^{-2} \). Hence, it becomes:\[ \left( \frac{1}{4} \right)^{-3/2} \]. According to exponent rules, \( (a^m)^n = a^{m \cdot n} \), this becomes \( 4^{3/2} \).

Step 3: Evaluate the Exponentiation

To compute \( 4^{3/2} \), rewrite the expression as \((2^2)^{3/2}\). Using exponent rules: \((a^m)^n = a^{m \cdot n}\), we get \(2^{2 \cdot \frac{3}{2}} = 2^3\). Hence, \( 2^3 = 8 \).

Key Concepts

Exponent Rules

Exponents are a shorthand way to express repeated multiplication of the same number. For example, \( a^n \) means "\( a \) multiplied by itself \( n \) times." There are several key rules for working with exponents that are vital for simplifying expressions correctly.

Here are some of the most important exponent rules:

• Multiplication Rule: When multiplying two exponents with the same base, add the exponents: \( a^m \times a^n = a^{m+n} \).
• Division Rule: When dividing two exponents with the same base, subtract the exponents: \( \frac{a^m}{a^n} = a^{m-n} \).
• Power of a Power Rule: When raising an exponent to another power, multiply the exponents: \( (a^m)^n = a^{m \cdot n} \).
• Negative Exponent Rule: A negative exponent represents the reciprocal of the base raised to the opposite positive exponent: \( a^{-n} = \frac{1}{a^n} \).

Understanding these rules is crucial for simplifying and evaluating expressions involving exponents accurately. In the given exercise, both the negative exponent and the power of a power rule are applied to find the final solution.

Simplifying Expressions

Simplifying mathematical expressions with exponents involves applying the appropriate exponent rules to reduce the complexity of the expression while maintaining its equivalence.

Let's look at the original exercise: \( \left(2^{-2}\right)^{-3/2} \).

**Step-by-Step Simplification**

• First Step: Begin with the innermost expression, \( 2^{-2} \). Using the negative exponent rule, rewrite it as \( \frac{1}{2^2} \).
• Second Step: Simplify by calculating \( \frac{1}{2^2} = \frac{1}{4} \).
• Third Step: Take the result, \( \frac{1}{4} \), and apply the outer exponent \( -\frac{3}{2} \). This transforms the expression into \( \left(\frac{1}{4}\right)^{-3/2} \).
• Fourth Step: Using the power of a power rule, this step becomes \( 4^{3/2} \).
• Fifth Step: Convert \( 4^{3/2} \) as \((2^2)^{3/2} \) to utilize easier calculation with a base of 2, solving \( 2^3 = 8 \).

By methodically applying the exponent rules, we simplify the complex expression into a manageable calculation, resulting in a solution of 8.

Negative Exponents

Negative exponents might look intimidating at first, but they simply indicate the reciprocal of a number raised to a positive exponent. It's essential to tackle negative exponents correctly to simplify expressions accurately.

For example, the expression \( a^{-n} \) can be rewritten as \( \frac{1}{a^n} \). This transformation makes it easier to handle calculations.

Consider the exercise again: \( 2^{-2} \) becomes \( \frac{1}{2^2} = \frac{1}{4} \). Negative exponents are not only used to simplify expressions but also bridge connections to other exponent rules, such as when combining or raising power to a power.

It’s critical to remember that:

• Negative exponents do not make the answer negative; they invert the base.
• Applying the reciprocal makes expressions easier to solve or further simplify.
• When both negative and positive exponents appear, address negative exponents first to eliminate their complexity.

Mastering negative exponents enhances your ability to tackle a wide range of algebraic expressions and mathematical challenges, ensuring solutions are precise and methodically sound.