Simplifying Algebraic Expressions With Fractional Exponents

Introduction

In this article, we will delve into simplifying a complex algebraic expression involving fractional exponents. The expression we aim to simplify is $(49 x^2 y)^{\frac{1}{4}}(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$. This problem tests our understanding of exponent rules and algebraic manipulation. Simplifying such expressions is a fundamental skill in mathematics, particularly in algebra and calculus. Mastery of these concepts enables us to tackle more complex problems in various fields, including physics, engineering, and computer science. By breaking down the problem step-by-step, we can clearly see how each rule of exponents is applied, enhancing our problem-solving abilities. The correct simplification will lead us to one of the provided answer choices, making this a great exercise in both algebraic manipulation and multiple-choice test-taking strategies. Let's embark on this journey to unravel the complexities of this expression and arrive at the correct simplified form.

Breaking Down the Expression

To effectively simplify the expression $(49 x^2 y)^{\frac{1}{4}}(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$, we will methodically apply the rules of exponents. Our first step involves distributing the outer exponents to each term within the parentheses. This is a crucial step because it allows us to work with simpler terms individually. Starting with the first part of the expression, $(49 x^2 y)^{\frac{1}{4}}$, we distribute the exponent $\frac{1}{4}$ to each factor: $49^{\frac{1}{4}}$, $(x^2)^{\frac{1}{4}}$, and $y^{\frac{1}{4}}$. Recognizing that $49$ is $7^2$, we can rewrite $49^{\frac{1}{4}}$ as $(7^2)^{\frac{1}{4}}$. Applying the power of a power rule, we multiply the exponents, resulting in $7^{\frac{2}{4}}$, which simplifies to $7^{\frac{1}{2}}$ or $\sqrt{7}$. For $(x^2)^{\frac{1}{4}}$, we again use the power of a power rule, multiplying the exponents to get $x^{\frac{2}{4}}$, which simplifies to $x^{\frac{1}{2}}$. The term $y^{\frac{1}{4}}$ remains as is since it is already in its simplest form. Moving on to the second part of the expression, $(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$, we first simplify $y^{\frac{2}{2}}$ to $y^1$ or simply $y$. Then we distribute the exponent $\frac{1}{3}$ to each factor: $27^{\frac{1}{3}}$, $(x^6)^{\frac{1}{3}}$, and $y^{\frac{1}{3}}$. We recognize that $27$ is $3^3$, so $27^{\frac{1}{3}}$ becomes $(3^3)^{\frac{1}{3}}$. Applying the power of a power rule, we multiply the exponents, resulting in $3^{\frac{3}{3}}$, which simplifies to $3^1$ or just $3$. For $(x^6)^{\frac{1}{3}}$, we multiply the exponents to get $x^{\frac{6}{3}}$, which simplifies to $x^2$. The term $y^{\frac{1}{3}}$ remains as is. By meticulously breaking down each part of the expression and applying the power of a power rule, we transform the initial complex form into a collection of simpler terms. This approach not only makes the problem more manageable but also minimizes the chances of errors in our calculations. In the next section, we will combine these simplified terms and further streamline the expression to arrive at the final answer.

Combining Simplified Terms

Having simplified the individual parts of the expression, we now focus on combining these terms to reach the final simplified form. From our previous steps, we have simplified $(49 x^2 y)^{\frac{1}{4}}$ to $\sqrt{7} x^{\frac{1}{2}} y^{\frac{1}{4}}$ and $(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$ to $3 x^2 y^{\frac{1}{3}}$. Our next step is to multiply these two simplified expressions together: $(\sqrt{7} x^{\frac{1}{2}} y^{\frac{1}{4}}) imes (3 x^2 y^{\frac{1}{3}})$. When multiplying terms with the same base, we add their exponents. We begin by multiplying the coefficients, which gives us $3\sqrt{7}$. Next, we multiply the $x$ terms: $x^{\frac{1}{2}} imes x^2$. Adding the exponents $\frac{1}{2}$ and $2$, we get $\frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}$, so the result is $x^{\frac{5}{2}}$. Then, we multiply the $y$ terms: $y^{\frac{1}{4}} imes y^{\frac{1}{3}}$. Adding the exponents $\frac{1}{4}$ and $\frac{1}{3}$, we need a common denominator, which is 12. Thus, $\frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}$, so the result is $y^{\frac{7}{12}}$. Combining all these results, we have $3\sqrt{7} x^{\frac{5}{2}} y^{\frac{7}{12}}$. Now, let's examine the answer choices provided. We need to determine which answer choice matches our simplified expression. It is crucial to carefully compare our result with the given options, ensuring that we have accounted for all terms and exponents correctly. If our expression does not directly match any of the provided choices, we may need to further simplify or rewrite our expression to make it comparable. This step is a critical part of the problem-solving process, as it requires attention to detail and a thorough understanding of algebraic manipulation. In the following section, we will compare our simplified expression with the provided options and select the correct answer.

Comparing with Answer Choices and Final Answer

After simplifying the expression $(49 x^2 y)^{\frac{1}{4}}(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$, we arrived at $3\sqrt{7} x^{\frac{5}{2}} y^{\frac{7}{12}}$. Now, we need to compare this result with the given answer choices to determine the correct option. The provided answer choices are:

A. $21 x^3 y$ B. $21 x^3 y^2$ C. $10 x^2 y$ D. $10 x^3 y$

Upon careful comparison, we notice that our simplified expression $3\sqrt{7} x^{\frac{5}{2}} y^{\frac{7}{12}}$ does not directly match any of the provided answer choices. This indicates that there might have been an error in our calculations or that the answer choices are not in their simplest form, or there may be a typo in the original question or answer choices. It's crucial to double-check each step of our simplification process to ensure accuracy. Let's revisit our steps:

1. We correctly distributed the exponents: $(49 x^2 y)^{\frac{1}{4}}$ became $\sqrt{7} x^{\frac{1}{2}} y^{\frac{1}{4}}$ and $(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$ became $3 x^2 y^{\frac{1}{3}}$.
2. We multiplied the coefficients: $\sqrt{7} imes 3 = 3\sqrt{7}$.
3. We added the exponents for $x$: $\frac{1}{2} + 2 = \frac{5}{2}$, resulting in $x^{\frac{5}{2}}$.
4. We added the exponents for $y$: $\frac{1}{4} + \frac{1}{3} = \frac{7}{12}$, resulting in $y^{\frac{7}{12}}$.

Our calculations appear to be correct. However, the absence of a matching answer choice suggests a potential issue with the question or the provided options. Given the discrepancy, it is important to acknowledge that none of the provided options accurately represent the simplified form of the given expression. In a real test scenario, this situation would warrant further investigation or clarification. For the purpose of this exercise, we must conclude that the correct simplified form is $3\sqrt{7} x^{\frac{5}{2}} y^{\frac{7}{12}}$, and none of the provided answer choices match this result.

Conclusion

In summary, we embarked on the task of simplifying the algebraic expression $(49 x^2 y)^{\frac{1}{4}}(27 x^6 y^{\frac{2}{2}})^{\frac{1}{3}}$. Through a step-by-step process, we applied the rules of exponents to break down the expression into manageable components. We distributed the fractional exponents, simplified terms, and combined like terms by adding their exponents. Our meticulous calculations led us to the simplified form $3\sqrt{7} x^{\frac{5}{2}} y^{\frac{7}{12}}$. However, upon comparing our result with the provided answer choices, we found no direct match. This discrepancy highlights the importance of careful calculation and attention to detail in mathematical problem-solving. While our derived simplified form is accurate based on the given expression, the absence of a matching answer choice suggests a potential issue with the question or the provided options. This situation underscores the need for critical evaluation and verification in mathematical exercises. Despite the absence of a matching answer, the process of simplifying the expression provided valuable practice in applying exponent rules and algebraic manipulation techniques. These skills are essential for success in more advanced mathematical topics and real-world applications. By diligently working through each step, we have reinforced our understanding of these fundamental concepts and enhanced our problem-solving abilities.