Problem 2 - A Graphical Approach to College Algebra

Question

Concept Check Match each expression from Group I with the correct choice from Group II. Choices may be used once, more than once, or not at all. (II) \(\begin{array}{lll}\text { A. } \frac{9}{4} & \text { B. }-\frac{9}{4}\end{array}\) \(\begin{array}{lll}\text { C. }-\frac{4}{9} & \text { D. } \frac{4}{9}\end{array}\) E. \(\frac{8}{27} \quad\) F. \(-\frac{27}{8}\) G. \(\frac{27}{8} \quad\) H. \(-\frac{8}{27}\) (I) $$\left(\frac{4}{9}\right)^{-3 / 2}$$

Step-by-Step Solution

Answer

The correct choice from Group II is G: \( \frac{27}{8} \).

Step 1: Understand the Expression in Group I

The expression given is \( \left(\frac{4}{9}\right)^{-3/2} \), which involves raising the fraction \( \frac{4}{9} \) to the power of \(-\frac{3}{2} \). The notation \( a^{-n} \) indicates the reciprocal, so we need to understand that this is equivalent to \( \frac{1}{(\frac{4}{9})^{3/2}} \).

Step 2: Simplifying the Exponent

The exponent \( \frac{3}{2} \) indicates both a cube and a square root operation. \( \left(\frac{4}{9}\right)^{3/2} \) is equivalent to \( \sqrt{\left(\frac{4}{9}\right)^3} \). Calculate \( \left(\frac{4}{9}\right)^3 \), which is \( \frac{4^3}{9^3} = \frac{64}{729} \).

Step 3: Calculate the Square Root

Calculate the square root of \( \frac{64}{729} \). The square root of \( 64 \) is \( 8 \) and the square root of \( 729 \) is \( 27 \), so \( \sqrt{\frac{64}{729}} = \frac{8}{27} \).

Step 4: Take the Reciprocal

Since the original problem involves \( \left(\frac{4}{9}\right)^{-3/2} \), or the reciprocal of the previous result, we must find \( \frac{1}{\frac{8}{27}} = \frac{27}{8} \).

Step 5: Match with Group II

Find the matching value in Group II. From Step 4, we determined that the solution is \( \frac{27}{8} \), which matches with option G in Group II.

Key Concepts

Reciprocal of a Fraction

When you encounter a negative exponent, such as in the expression \( a^{-n} \), it hints at the concept of a reciprocal. Here, the negative sign signifies the inverse of the base fraction. So, \( \left( \frac{4}{9} \right)^{-3/2} \) becomes \( \frac{1}{\left( \frac{4}{9} \right)^{3/2}} \).
This transformation is pivotal because it turns the problem into a more straightforward multiplication process. Instead of directly computing a complex power with a negative exponent, we first deal with the positive exponent and then find the reciprocal.
In essence, the reciprocal of a fraction \( \frac{a}{b} \) is simply \( \frac{b}{a} \). Thus, understanding this fundamental concept allows you to efficiently handle and solve problems involving negative exponents.

Cube and Square Root Operations

Handling fractional exponents, such as \( \frac{3}{2} \), involves both rooting and powering operations. This is because the numerator and denominator in the exponent represent different tasks. The denominator (2, in this case) corresponds to a square root operation, while the numerator (3) corresponds to cube or power operations.
In the expression \( \left( \frac{4}{9} \right)^{3/2} \), you first cube the fraction \( \frac{4}{9} \), obtaining \( \left( \frac{4}{9} \right)^3 = \frac{64}{729} \).
Next, find the square root: \( \sqrt{\frac{64}{729}} \). This involves finding the square root of both the numerator and the denominator separately: \( \sqrt{64} = 8 \) and \( \sqrt{729} = 27 \). Therefore, the result of the square root operation is \( \frac{8}{27} \).
Mastering these root and power operations is crucial for understanding fractional exponents and simplifying expressions appropriately.

Exponentiation Rules

Understanding exponentiation rules gives you a strong mathematical foundation to solve expressions like \( \left( \frac{4}{9} \right)^{-3/2} \). To approach such expressions, grasp these essential rules:

• A negative exponent, such as \( a^{-n} \), indicates taking the reciprocal of the base and using a positive exponent: \( \frac{1}{a^n} \).
• Fractional exponents like \( \frac{m}{n} \) mean that you root and then power, or power and then root. This duality assists in simplifying the expression.
• An exponent of 1/2 signifies a square root, 1/3 a cube root, and so on. Multiply by these fractional values to determine roots or powers.

By following these rules, you can manipulate such exponent expressions effectively, breaking them down into more manageable parts. Whether dealing with negative, fractional, or whole exponents, understanding these principles allows for accurate computation and matching of expressions.