Problem 74 - College Algebra with Modeling and Visualization

Question

Evaluate the expression by band. Approximate the answer to the nearest hundredth when appropriate. $$ \left(\frac{16}{25}\right)^{-3 / 2} $$

Step-by-Step Solution

Answer

The expression evaluates to approximately 1.95.

Step 1: Understand the Negative Exponent

The expression \( \left(\frac{16}{25}\right)^{-3/2} \) has a negative exponent. A negative exponent indicates the reciprocal of the base raised to the positive exponent. Therefore, we rewrite it as \( \left(\frac{25}{16}\right)^{3/2} \).

Step 2: Apply the Fractional Exponent

The exponent \( \frac{3}{2} \) can be split as \( \left(\left(\frac{25}{16}\right)^{1/2}\right)^3 \). First, we calculate the square root \( \left(\frac{25}{16}\right)^{1/2} \), which is \( \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4} \).

Step 3: Raise to the Power of Three

After finding the square root, we raise the result to the power of 3: \( \left(\frac{5}{4}\right)^3 = \frac{5^3}{4^3} = \frac{125}{64} \).

Step 4: Convert to Decimal and Approximate

Finally, convert \( \frac{125}{64} \) to a decimal by performing the division: \( \frac{125}{64} = 1.953125 \). Approximating this value to the nearest hundredth gives us \( 1.95 \).

Key Concepts

Negative Exponents

Negative exponents might seem tricky at first, but they're quite simple once you get the hang of them. A negative exponent tells us to take the reciprocal of the base and then apply the positive version of the exponent. For instance, with our expression \( \left(\frac{16}{25}\right)^{-3/2} \), the negative exponent \(-3/2\) initially suggests inverting the fraction. We flip the base to \( \frac{25}{16} \), turning the negative exponent into a positive one as \( \left(\frac{25}{16}\right)^{3/2} \).

• Remember: \( a^{-b} = \frac{1}{a^b} \)
• This rule applies universally, regardless of whether the base is a fraction or a whole number.

Once you invert the base, the problem becomes straightforward, allowing you to handle the exponentiation without any negativity involved.

Fractional Exponents

Fractional exponents can be seen as a bridge between exponents and roots, providing an alternative notation. In the case of \( \frac{3}{2} \), the numerator 3 specifies a third power, while the denominator 2 denotes a square root. Thus, \( \left(\frac{25}{16}\right)^{3/2} \) breaks down into two steps: first, compute the square root \( \left(\frac{25}{16}\right)^{1/2} \), and then cube the result.

• Break it into parts: \( \left(a^b\right)^c = a^{bc} \)
• The root comes first, then the power: \( \left(\text{root}\right) = \left(\text{power}\right) \)

By approaching fractional exponents in this manner, they are less daunting and much easier to manage and solve.

Rational Exponents

Understanding rational exponents is crucial because they reveal a deeper relationship between powers and roots. Rational exponents, like \( \frac{3}{2} \), imply multiple operations: in essence, performing a root and a power simultaneously. Here, \( \left(\frac{25}{16}\right)^{3/2} \) guides us first to take the square root (denominator of 2) and then raise it to the power of 3 (numerator of 3).

• The denominator indicates the type of root.
• The numerator specifies the exponent you apply after taking the root.

Rational exponents offer a compact way to represent these operations, simplifying the process by providing a structured method to manipulate both radicals and powers in one neat package.

Radicals

Radicals function as an essential part of mathematical operations, mainly since they represent roots. In our example, calculating \( \left(\frac{25}{16}\right)^{1/2} \) means finding the square root of the fraction. By simplifying each part individually, we find \( \frac{\sqrt{25}}{\sqrt{16}} = \frac{5}{4} \).

• Always simplify the numerator and denominator separately when possible.
• Fast checking: Look for perfect squares, as they simplify quickly under a root.

Understanding how and when to apply radicals is key to solving complex expressions. Once you grasp their function, radicals become a powerful tool in a mathematician's kit, helping to solve equations efficiently and accurately.