Simplifying Expressions With Fractional Exponents

In the world of mathematics, simplifying expressions is a fundamental skill. It allows us to take complex equations and reduce them to their most basic form, making them easier to understand and work with. This is especially important when dealing with exponents, particularly fractional exponents. In this article, we will dive deep into the process of simplifying expressions involving fractional exponents, providing you with a comprehensive understanding of the rules and techniques involved. Let's break down the expression $(32b^{\frac{15}{2}}y^{\frac{15}{2}})^{\frac{6}{5}}$ step-by-step, ensuring clarity and thoroughness in our explanation. The goal is to transform the given expression into its simplest equivalent form, making it more manageable and revealing its underlying structure.

Understanding Fractional Exponents

Before we tackle the given expression, let's quickly recap what fractional exponents mean. A fractional exponent like $\frac{a}{b}$ represents both a power and a root. The numerator (a) indicates the power to which the base is raised, while the denominator (b) indicates the root to be taken. For example, $x^{\frac{1}{2}}$ is the square root of x, and $x^{\frac{2}{3}}$ is the cube root of $x^2$. This understanding is crucial because fractional exponents are the heart of the expression we are going to simplify. Grasping this concept allows us to convert between radical and exponential forms, a skill that will prove invaluable in simplifying complex expressions. Remember, the laws of exponents still apply when exponents are fractions, which means we can multiply exponents when raising a power to a power, and we can add exponents when multiplying like bases. Let’s keep this in mind as we move forward. The ability to manipulate fractional exponents effectively is not just about following rules; it's about understanding the relationship between exponents and radicals, and how they interact with each other.

Breaking Down the Expression: $(32b^{\frac{15}{2}}y^{\frac{15}{2}})^{\frac{6}{5}}$

Our task is to simplify $(32b^{\frac{15}{2}}y^{\frac{15}{2}})^{\frac{6}{5}}$. The first step in simplifying this expression is to apply the power of a product rule. This rule states that $(ab)^n = a^n b^n$. In our case, we have a product inside the parentheses raised to the power of $\frac{6}{5}$. Applying the rule, we get:

$(32b^{\frac{15}{2}}y^{\frac{15}{2}})^{\frac{6}{5}} = 32^{\frac{6}{5}} imes (b^{\frac{15}{2}})^{\frac{6}{5}} imes (y^{\frac{15}{2}})^{\frac{6}{5}}$

This step is vital because it separates the expression into more manageable parts. We have now distributed the exponent $\frac{6}{5}$ to each factor within the parentheses: the constant 32, the variable b raised to the power of $\frac{15}{2}$, and the variable y also raised to the power of $\frac{15}{2}$. This distribution allows us to deal with each part individually, making the simplification process much clearer. Remember, the key to simplifying complex expressions often lies in breaking them down into smaller, more manageable components. By applying the power of a product rule, we’ve taken the first step towards simplifying our expression. Understanding the rationale behind each step is as important as the steps themselves.

Simplifying $32^{\frac{6}{5}}$

Next, let's simplify $32^{\frac{6}{5}}$. We can rewrite 32 as $2^5$, so our expression becomes $(2^5)^{\frac{6}{5}}$. Now we use the power of a power rule, which states that $(a^m)^n = a^{mn}$. Multiplying the exponents, we get:

$(2^5)^{\frac{6}{5}} = 2^{5 imes \frac{6}{5}} = 2^6$

And $2^6$ is simply 64. This demonstrates how converting a number into its prime factorization can greatly simplify expressions with fractional exponents. Recognizing that 32 is a power of 2 allows us to easily apply the power of a power rule, turning a potentially complex calculation into a straightforward one. This technique is not only useful for constants but can also be applied to variables with exponents. The ability to identify and utilize such relationships is a hallmark of strong mathematical proficiency. Remember, simplifying expressions is not just about finding the right answer; it's about finding the most efficient and elegant path to that answer. This step highlights the importance of recognizing patterns and applying the appropriate exponent rules.

Simplifying $(b^{\frac{15}{2}})^{\frac{6}{5}}$ and $(y^{\frac{15}{2}})^{\frac{6}{5}}$

Now, let's focus on the variable terms. We have $(b^{\frac{15}{2}})^{\frac{6}{5}}$ and $(y^{\frac{15}{2}})^{\frac{6}{5}}$. Again, we apply the power of a power rule: $(a^m)^n = a^{mn}$. For the b term, we get:

$(b^{\frac{15}{2}})^{\frac{6}{5}} = b^{\frac{15}{2} imes \frac{6}{5}} = b^{\frac{15 imes 6}{2 imes 5}} = b^{\frac{90}{10}} = b^9$

Similarly, for the y term, we have:

$(y^{\frac{15}{2}})^{\frac{6}{5}} = y^{\frac{15}{2} imes \frac{6}{5}} = y^{\frac{15 imes 6}{2 imes 5}} = y^{\frac{90}{10}} = y^9$

This step demonstrates the power of the power rule in simplifying expressions with nested exponents. By multiplying the exponents, we’ve transformed fractional exponents into whole numbers, making the expression significantly simpler. Notice how the multiplication of fractions simplifies to a whole number exponent, which is a common outcome when dealing with expressions of this nature. This process not only simplifies the expression but also reveals the underlying structure and relationships between the variables. Remember, the goal is not just to arrive at the correct answer but to understand the mathematical principles that allow us to do so. This understanding is what enables us to tackle more complex problems with confidence. The consistent application of exponent rules is key to mastering these types of simplifications.

Putting It All Together

Now we combine the simplified terms. We found that $32^{\frac{6}{5}} = 64$, $(b^{\frac{15}{2}})^{\frac{6}{5}} = b^9$, and $(y^{\frac{15}{2}})^{\frac{6}{5}} = y^9$. Multiplying these together, we get:

$64b^9y^9$

This is the fully simplified form of the original expression. By breaking down the problem into smaller steps and applying the rules of exponents systematically, we were able to transform a seemingly complex expression into a much simpler one. This final step showcases the beauty of mathematical simplification: taking a complex expression and reducing it to its most elegant and understandable form. The ability to synthesize the results of individual simplifications into a cohesive whole is a critical skill in mathematics. It’s not enough to be able to perform individual operations; you must also be able to see how they fit together to solve the overall problem. This comprehensive approach to problem-solving is what distinguishes a true understanding of mathematical concepts.

Final Answer

Therefore, the simplified form of $(32b^{\frac{15}{2}}y^{\frac{15}{2}})^{\frac{6}{5}}$ is $64b^9y^9$. This exercise highlights the importance of understanding and applying the rules of exponents, particularly when dealing with fractional exponents. Remember, simplifying expressions is not just about getting the right answer; it's about understanding the process and the underlying mathematical principles. By mastering these techniques, you’ll be well-equipped to tackle more complex mathematical problems. Keep practicing and exploring different types of expressions, and you’ll find that simplifying becomes second nature. The journey of mathematical discovery is filled with challenges, but the rewards of understanding and mastery are well worth the effort.

In conclusion, understanding fractional exponents and the rules governing them allows us to simplify complex expressions effectively. By breaking down the problem into smaller, manageable steps, we can apply the power of a product rule and the power of a power rule to arrive at the simplified form. This process not only provides the correct answer but also reinforces our understanding of mathematical principles. For further exploration of exponent rules and simplification techniques, consider visiting resources like Khan Academy's Algebra I section on exponents.