Problem 1 - 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

Match Group I expression with choice E: \(\frac{8}{27}\).

Step 1: Understand the Expression

The expression given in Group I is \( \left(\frac{4}{9}\right)^{3/2} \). This means we need to raise the fraction \( \frac{4}{9} \) to the power of \( \frac{3}{2} \).

Step 2: Interpret the Exponent

The exponent \( \frac{3}{2} \) can be broken down into two parts: the numerator \( 3 \) and the denominator \( 2 \). This can be interpreted as the square root of the fraction raised to the power of 3.

Step 3: Calculate the Square Root

Calculate the square root of \( \frac{4}{9} \): \[ \sqrt{\frac{4}{9}} = \frac{2}{3} \].

Step 4: Apply the Power of 3

Raise the result from Step 3 to the power of 3: \[ \left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27} \].

Step 5: Match with Group II

Now, check which option in Group II matches \( \frac{8}{27} \). Option E corresponds to \( \frac{8}{27} \).

Key Concepts

Rational Exponents

Rational exponents are a way to express powers and roots together in a succinct form. A rational exponent like \( \frac{3}{2} \) indicates two operations: root extraction and exponentiation. The denominator indicates the root, and the numerator indicates the power.
When you see \( x^{\frac{m}{n}} \), it instructs you to take the \( n \)-th root of \( x \) and then raise that result to the \( m \)-th power. Alternatively, you can raise \( x \) to the \( m \)-th power first and then take the \( n \)-th root. Both methods lead to the same result.

• Denominator of \( \frac{3}{2} \) says "take the square root."
• Numerator of \( \frac{3}{2} \) says "raise to the third power."

This process might seem complex at first glance, but breaking it into two clear, distinct steps aids the understanding.

Fractional Exponents

Fractional exponents can initially be confusing, but they offer a convenient way to express both roots and powers in a uniform notation. If we consider the base of a fractional exponent, the number underneath signifies the root. Meanwhile, the numerator indicates how many times we multiply the base with itself after extracting the root.
For example, let's break down \( \left( \frac{4}{9} \right)^{\frac{3}{2}} \):

• Firstly, the fraction \( \frac{4}{9} \) is squared rooted to become \( \frac{2}{3} \).
• Subsequently, \( \frac{2}{3} \) is raised to the third power, yielding \( \frac{8}{27} \).

Using fractional exponents can greatly simplify complex calculations and transform them into more manageable sequences of operations.

Mathematical Matching Problems

Mathematical matching problems involve pairing expressions from one group with the equivalently transformed expressions in another group. This type of problem is an effective way to test understanding, as it encourages recognizing relationships between different mathematical forms.
To excel in matching problems,

• Carefully analyze each expression in the context of any mathematical rules or properties.
• Use step-by-step transformations when necessary to see which expression from the other group it matches.

For instance, by transforming \( \left( \frac{4}{9} \right)^{\frac{3}{2}} \) into \( \frac{8}{27} \) through the correct application of fractional exponent rules, you ensure a correct match from Group I to Group II. Over time, repeated practice with these problems improves both perception of mathematical connections and overall problem-solving skills.