Problem 6 - 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) $$\left(\frac{8}{27}\right)^{-2 / 3}$$

Step-by-Step Solution

Answer

(I) matches with A: \( \frac{9}{4} \).

Step 1: Recall the Exponent Rule for Roots

The expression \( \left( \frac{8}{27} \right)^{-2/3} \) indicates that we need to deal with both a negative exponent and a fractional exponent. The rule for fractional exponents is: \( a^{m/n} = \sqrt[n]{a^m} \). Here, the exponent \(-2/3\) suggests taking the cube root, then squaring the result, and finally taking the reciprocal due to the negative exponent.

Step 2: Compute the Cube Root

First, find the cube root of \( \frac{8}{27} \). This is \( \sqrt[3]{8} \) over \( \sqrt[3]{27} \), which simplifies to \( \frac{2}{3} \) because \( \sqrt[3]{8} = 2 \) and \( \sqrt[3]{27} = 3 \).

Step 3: Square the Result

Next, square the result from Step 2: \( \left( \frac{2}{3} \right)^2 = \frac{4}{9} \).

Step 4: Apply the Negative Exponent

The negative exponent indicates that we should take the reciprocal of the result from Step 3. The reciprocal of \( \frac{4}{9} \) is \( \frac{9}{4} \).

Step 5: Match with Group II

Now, match the computed value \( \frac{9}{4} \) with the choices in Group II. The correct match is A. \( \frac{9}{4} \).

Key Concepts

Exponent Rules

When dealing with expressions like \( \left( \frac{8}{27} \right)^{-2/3} \), a good grasp of exponent rules is essential. Exponent rules are a set of guidelines used to simplify expressions involving powers of the same base. They help us manage how to handle powers more effectively.

One important rule for fractional exponents gives us the formula: \( a^{m/n} = \sqrt[n]{a^m} \). This rule tells us that a fractional exponent such as \( m/n \) means we raise the base \( a \) to the \( m \, \text{th} \) power and then take the \( n \, \text{th} \) root. Fractional exponents allow us to express roots and powers in the same notation.

• Use this rule with care to simplify expressions.
• An exponent in the form \( -m/n \) also indicates a reciprocal operation, but we'll explore that in a bit!

Cube Root

A cube root is a special case of fractional exponents where the exponent is \( 1/3 \). It answers the question ‘What number, when multiplied by itself three times, gives me the original number?’ It's related to the idea of reversing a cube power operation.

To compute the cube root of a fraction like \( \frac{8}{27} \), we separately take the cube root of both the numerator and the denominator.

• \( \sqrt[3]{8} = 2 \) because \( 2 \times 2 \times 2 = 8 \)
• \( \sqrt[3]{27} = 3 \) because \( 3 \times 3 \times 3 = 27 \)

Thus, \( \sqrt[3]{\frac{8}{27}} = \frac{2}{3} \). Remember, calculating roots for each part of a fraction separately simplifies complex expressions.

Reciprocal

The concept of a reciprocal can be a little tricky but is straightforward once you understand it. The reciprocal of a fraction simply means flipping the numerator and the denominator. For example, the reciprocal of \( \frac{4}{9} \) is \( \frac{9}{4} \).

In the context of exponents, when you encounter a negative exponent such as in \( a^{-b} \), it indicates a reciprocal action. Therefore, \( a^{-b} = \frac{1}{a^b} \).

• This rule is crucial for simplifying equations involving negative exponents.
• After computing an expression with a negative exponent, simply find the reciprocal – reverse the fraction!

Remembering these rules will help you tackle any fraction or exponent challenge at ease.