Problem 19 - Algebra and Trigonometry

Question

\(15-24\) . Evaluate each expression. $$ \begin{array}{llll}{\text { (a) }\left(\frac{4}{9}\right)^{-1 / 2}} & {\text { (b) }(-32)^{2 / 5}} & {\text { (c) }(-125)^{-1 / 3}}\end{array} $$

Step-by-Step Solution

Answer

(a) \(\frac{3}{2}\), (b) 4, (c) \(-\frac{1}{5}\).

Step 1: Evaluate Expression (a)

To evaluate \( \left(\frac{4}{9}\right)^{-1/2} \), first remember that a negative exponent indicates a reciprocal. Thus, \( \left(\frac{4}{9}\right)^{-1/2} = \left(\frac{9}{4}\right)^{1/2} \). Next, compute the square root. The square root of 9 is 3, and the square root of 4 is 2. Therefore, \( \left(\frac{9}{4}\right)^{1/2} = \frac{3}{2} \).

Step 2: Evaluate Expression (b)

For \( (-32)^{2/5} \), we need to find the fifth root (denominator 5) and then square the answer (numerator 2). The fifth root of -32 is -2, because \((-2)^5 = -32\). Now, square -2 to get \((-2)^2 = 4\). Thus, \((-32)^{2/5} = 4\).

Step 3: Evaluate Expression (c)

To evaluate \((-125)^{-1/3}\), recognize that the negative exponent means we will take the reciprocal after computing the cube root (since the exponent is \(-1/3\)). First, find the cube root of -125, which is -5, because \((-5)^3 = -125\). The reciprocal of -5 is \(-\frac{1}{5}\). Therefore, \((-125)^{-1/3} = -\frac{1}{5}\).

Key Concepts

Negative Exponents

Negative exponents might look confusing, but they actually have a simple meaning. When you see a number raised to a negative exponent, it means you're looking at the reciprocal of that number with a positive exponent.

• For instance, let's consider the expression \(a^{-n}\). The negative exponent \(-n\) tells us to invert \(a\), i.e., take its reciprocal, and then raise the result to the positive \(n\) exponent: \(a^{-n} = \frac{1}{a^n}\).
• In our example \(\left(\frac{4}{9}\right)^{-1/2}\), the negative exponent \(-1/2\) implies taking the reciprocal of \(\frac{4}{9}\) first, resulting in \(\left(\frac{9}{4}\right)^{1/2}\).

Practicing with negative exponents can make it easier to understand how they work, shifting between transformations of numbers and operations.

Fractional Exponents

Fractional exponents, also known as rational exponents, involve both roots and powers. They can stand for roots, similar to a square or cube root, using the fraction's denominator, while the numerator indicates raising to a power.

• For example, an expression like \(a^{m/n}\) means you take the nth root of \(a\) and raise the result to the power of \(m\): \(a^{m/n} = (\sqrt[n]{a})^m\).
• In step b of our solution, \((-32)^{2/5}\) translates to taking the fifth root of -32 and then squaring the answer. The fifth root of \(-32\) is \(-2\), so the entire expression becomes \((-2)^2\), which results in \(4\).

Using fractional exponents provides a concise way to denote complex expressions with both roots and powers.

Reciprocal of a Number

The reciprocal of a number is quite simply what you multiply the original number by to get one. In mathematical terms, for a given number \(a\), its reciprocal is \(\frac{1}{a}\). This concept is frequently used in dealing with negative exponents or to solve fractional expressions.

• To find the reciprocal of a fraction, simply invert it. For the fraction \(\frac{a}{b}\), the reciprocal would be \(\frac{b}{a}\).
• In step c of our solution, the expression \((-125)^{-1/3}\) involves first finding the cube root of \(-125\), which is \(-5\). The negative exponent then requires taking the reciprocal, resulting in \(-\frac{1}{5}\).

Understanding reciprocals is essential for solving equations involving fractions and negative exponents.

Cube Roots

Cube roots find the original number that, when multiplied by itself three times, results in the given number. The cube root is denoted as \(\sqrt[3]{a}\) or \(a^{1/3}\).

• If you have \(x^3 = a\), then \(x\) is the cube root of \(a\).
• When evaluating expressions like \((-125)^{-1/3}\), finding the cube root provides the starting point for manipulation. In this case, the cube root of \(-125\) is \(-5\), because \((-5)^3 = -125\).

Cube roots are particularly important in complex problems involving powers and roots, helping to break down larger expressions into manageable parts.