Simplify (y^(4/3) * Y^(2/3))^(-1/2): A Step-by-Step Guide

When you're faced with a complex-looking mathematical expression like $\left(y^{\frac{4}{3}} \cdot y^{\frac{2}{3}}\right)^{-\frac{1}{2}}$, it's easy to feel a bit overwhelmed. But don't worry! With a few key rules of exponents, we can break it down and simplify it to something much more manageable. The power of a power property is going to be our best friend here, alongside another handy rule for combining terms with the same base. Let's dive into how we can tackle this problem and arrive at the correct answer, which turns out to be surprisingly simple.

Understanding the Power of a Power Property

The power of a power property in mathematics is a fundamental rule that helps us simplify expressions involving exponents. It states that when you raise a power to another power, you multiply the exponents. In formulaic terms, this looks like: $(a^m)^n = a^{m \cdot n}$. So, if you have something like $(x^3)^2$, you simply multiply the exponents 3 and 2 to get $x^{3 \cdot 2} = x^6$. This property is incredibly useful for tidying up expressions that might otherwise look quite complicated. It's one of the building blocks for understanding more advanced algebraic manipulations, and mastering it will make tackling other exponent problems a breeze. Imagine you have a term that's already an exponent, like $y^2$, and you need to raise that entire term to another power, say $-1/2$. Applying the power of a power property means we take the existing exponent (2) and multiply it by the new exponent (-1/2). This leads to $y^{2 \cdot (-\frac{1}{2})}$, which simplifies directly to $y^{-1}$. This rule is the key to unlocking the simplification of our main expression, and it's worth remembering its straightforward application.

Simplifying the Expression Inside the Parentheses

Before we can apply the power of a power property to the entire expression, we need to deal with what's happening inside the parentheses: $\left(y^{\frac{4}{3}} \cdot y^{\frac{2}{3}}\right)$. Here, we have two terms with the same base, 'y', being multiplied together. The rule for multiplying exponents with the same base is to add their exponents: $a^m \cdot a^n = a^{m+n}$. So, for our expression, we add the exponents $\frac{4}{3}$ and $\frac{2}{3}$. The addition is straightforward because they already have a common denominator: $\frac{4}{3} + \frac{2}{3} = \frac{4+2}{3} = \frac{6}{3}$. This fraction simplifies further to just 2. Therefore, the expression inside the parentheses, $y^{\frac{4}{3}} \cdot y^{\frac{2}{3}}$, simplifies to $y^2$. This step is crucial because it reduces the complexity of the problem significantly, setting us up perfectly for the next stage of simplification. We've taken a product of powers and turned it into a single power, making the subsequent steps much cleaner.

Applying the Power of a Power Property

Now that we've simplified the inside of the parentheses to $y^2$, our expression becomes $\left(y^2\right)^{-\frac{1}{2}}$. This is precisely where the power of a power property shines! As we discussed earlier, when you raise a power to another power, you multiply the exponents. In this case, we have the base 'y' raised to the power of 2, and that entire term is raised to the power of $-\frac{1}{2}$. So, we multiply the exponents: $2 \cdot \left(-\frac{1}{2}\right)$. The calculation $2 \times -\frac{1}{2}$ equals $-1$. This means our expression simplifies to $y^{-1}$. This is a significant reduction from the original form, demonstrating the efficiency of these exponent rules. The direct application of the power of a power property here transforms the expression into its most basic exponential form, making the final step extremely clear.

Understanding Negative Exponents

We've simplified the expression down to $y^{-1}$. Now, we need to understand what a negative exponent means. The rule for negative exponents is that $a^{-n} = \frac{1}{a^n}$. In our case, we have $y^{-1}$, which means 'y' raised to the power of -1. Applying the rule, we get $\frac{1}{y^1}$. Since any number raised to the power of 1 is just itself, $y^1$ is simply 'y'. Therefore, $y^{-1}$ simplifies to $\frac{1}{y}$. This final step converts the expression from having a negative exponent to a form that is often considered more standard or simplified in many mathematical contexts. Understanding this rule is key to presenting the final answer in its most recognizable form and completing the simplification process.

Final Answer and Options

After applying the rules of exponents, we found that $\left(y^{\frac{4}{3}} \cdot y^{\frac{2}{3}}\right)^{-\frac{1}{2}}$ simplifies to $\frac{1}{y}$. Now, let's look at the options provided:

A. y B. $y^{\frac{1}{2}}$ C. $\frac{1}{y}$

Our calculated simplified form, $\frac{1}{y}$, directly matches option C. This confirms that our step-by-step application of exponent rules, including combining terms with the same base, using the power of a power property, and understanding negative exponents, has led us to the correct answer. It's always satisfying when the result aligns perfectly with one of the choices!

Conclusion

Simplifying expressions involving exponents might seem daunting at first, but by systematically applying the properties of exponents, it becomes a manageable and even enjoyable process. We've seen how the rule for multiplying terms with the same base ($a^m \cdot a^n = a^{m+n}$) allowed us to simplify the expression inside the parentheses. Then, the power of a power property ($(a^m)^n = a^{m \cdot n}$) was instrumental in reducing the overall complexity. Finally, understanding how to handle negative exponents ($a^{-n} = \frac{1}{a^n}$) helped us arrive at the final, clean answer. This journey from $\left(y^{\frac{4}{3}} \cdot y^{\frac{2}{3}}\right)^{-\frac{1}{2}}$ to $\frac{1}{y}$ showcases the elegance and power of mathematical rules. Remember these properties, and you'll be well-equipped to tackle similar problems.

For further exploration of exponent rules and their applications, you can visit Khan Academy's excellent resources on algebra and exponents. They offer detailed explanations and practice problems that can reinforce your understanding.