91Ó°ÊÓ

Negative exponents might seem puzzling at first, but they follow a simple rule: any base with a negative exponent can be converted into a reciprocal with a positive exponent. This rule is essential when simplifying expressions. For example:

• Given a term like \(x^{-a}\), it can really be seen as \(\frac{1}{x^a}\).
• This means that instead of dealing with negative powers, you can rewrite them as fractions.

Let's look at a practical example. In the step-by-step solution, the term \(y^{-1/5}\) was rewritten as \(\frac{1}{y^{1/5}}\). This transformation makes it easier to work with and understand the expression, as you no longer have to think in terms of negative exponents. Instead, you simplify by placing the term with a negative exponent in the denominator of a fraction.

The distributive property is a fundamental concept in algebra that helps us simplify complex expressions. It says that a single term multiplied by terms inside parentheses should be distributed, or applied, to each term independently. This is often expressed as:

• \(a(b + c) = ab + ac\)

But when it comes to dealing with exponents, the distributive property helps in another way: distributing the exponent to each factor within the brackets. For instance:

• Given \((xy)^n\), you would treat it as \(x^n \cdot y^n\).

In the example provided, the expression \((x^{-5} y^{1/3})^{-3/5}\) required distributing the power of \(-\frac{3}{5}\) to both \(x^{-5}\) and \(y^{1/3}\). This step leads to breaking down the problem into simpler steps, allowing for individual handling of each term under the power.

Fractional exponents might seem complex, but they're easy to understand once you know the basics. They represent roots and powers all in one. Here's how:

• A fractional exponent like \(a^{m/n}\) can be rewritten as the nth root of a raised to the power of m: \(\sqrt[n]{a^m}\).

This means you're handling both the root and the exponent in the same move. In the step-by-step example, the fractional exponents were distributed accordingly. For instance, in \((4r^8s^{-1/2})^{1/2}(32s^{-5/4})^{-1/5}\), applying the rules of fractional exponents:

• \((4^{1/2})\) becomes \(2\), because the square root of 4 is 2.
• The term \(32^{-1/5}\) simplifies to \(\frac{1}{2}\), reflecting the idea of the 5th root.

Understanding and employing these basics help in transforming complex expressions into much simpler forms, making mathematical problems much less intimidating.