Finding The Value Of K In The Equation X^k Y^4(2x^3 + 7x^2y^4) = 2x^4y^4 + 7x^3y^8

Jul 15, 2025 by ADMIN 83 views

Introduction

In this article, we will delve into the realm of algebraic equations and explore how to determine the unknown exponent that satisfies a given equation. Our focus will be on the equation x^k y^4(2x3 + 7x^2y4) = 2x^4y4 + 7x^3y8, where we aim to find the value of k that makes the statement true. This problem involves understanding the rules of exponents and applying them strategically to simplify the equation and isolate the variable k. By carefully examining the equation's structure and performing algebraic manipulations, we can successfully unravel the mystery of k and gain a deeper appreciation for the elegance of mathematical problem-solving.

Understanding the Equation

Before we jump into solving for k, let's first dissect the equation and understand its components. The equation x^k y^4(2x3 + 7x^2y4) = 2x^4y4 + 7x^3y8 involves variables x and y, along with an unknown exponent k. The left-hand side of the equation is a product of x^k y^4 and the binomial (2x^3 + 7x^2y4), while the right-hand side is the sum of two terms: 2x^4y4 and 7x^3y8. Our goal is to manipulate the equation algebraically to isolate k and determine its value. This process will involve applying the distributive property, simplifying expressions with exponents, and comparing terms on both sides of the equation.

To effectively tackle this problem, it's crucial to have a solid grasp of the fundamental rules of exponents. Specifically, we'll be using the following properties:

Product of Powers: x^m * xⁿ = x^m+n
Power of a Product: (xy)ⁿ = xⁿyⁿ

With these principles in mind, we can embark on the journey of solving for k with confidence and precision. The equation might seem daunting at first glance, but by breaking it down into smaller, manageable steps, we can systematically uncover the value of the unknown exponent and appreciate the beauty of algebraic manipulation.

Applying the Distributive Property

Our first step in solving for k is to apply the distributive property to the left-hand side of the equation. The distributive property states that a(b + c) = ab + ac. Applying this property to our equation, x^k y^4(2x3 + 7x^2y4) = 2x^4y4 + 7x^3y8, we multiply x^k y^4 by each term inside the parentheses:

x^ky⁴ * 2x³ + x^ky⁴ * 7x²y⁴ = 2x⁴y⁴ + 7x³y⁸

This gives us:

2x^kx³y⁴ + 7x^kx²y⁴y⁴ = 2x⁴y⁴ + 7x³y⁸

Now, we can simplify the terms by using the product of powers rule, which states that x^m * xⁿ = x^m+n. Applying this rule to the exponents of x and y in each term, we get:

2x^k+3y⁴ + 7x^k+2y⁸ = 2x⁴y⁴ + 7x³y⁸

This simplified equation is now in a form where we can directly compare the terms on both sides and deduce the value of k. By carefully matching the exponents of x and y, we can create equations that will allow us to solve for the unknown variable. This step is crucial in unraveling the mystery of k and bringing us closer to the final solution. The application of the distributive property has transformed the equation into a more manageable form, setting the stage for the next step in our algebraic journey.

Comparing Terms and Solving for k

Now that we have simplified the equation to 2x^k+3y⁴ + 7x^k+2y⁸ = 2x⁴y⁴ + 7x³y⁸, we can compare the terms on both sides to determine the value of k. To do this, we'll match the terms with the same variables and exponents. Notice that we have two terms on each side of the equation, each involving x and y raised to certain powers.

Let's first compare the terms with y⁴. On the left-hand side, we have 2x^k+3y⁴, and on the right-hand side, we have 2x⁴y⁴. For these terms to be equal, the exponents of both x and y must be the same. Since the exponents of y are already equal (both are 4), we can focus on the exponents of x. This gives us the equation:

k + 3 = 4

Solving for k, we subtract 3 from both sides:

k = 4 - 3

k = 1

Now, let's verify this value of k by comparing the terms with y⁸. On the left-hand side, we have 7x^k+2y⁸, and on the right-hand side, we have 7x³y⁸. Again, for these terms to be equal, the exponents of x must be the same. This gives us the equation:

k + 2 = 3

Solving for k, we subtract 2 from both sides:

k = 3 - 2

k = 1

As we can see, both pairs of terms give us the same value for k, which is 1. This confirms that our solution is consistent and correct. By carefully comparing the terms and solving the resulting equations, we have successfully determined the value of k that makes the original statement true. This process highlights the power of algebraic manipulation and the importance of paying close attention to the details of the equation.

Solution

By carefully applying the distributive property and comparing terms on both sides of the equation x^k y^4(2x3 + 7x^2y4) = 2x^4y4 + 7x^3y8, we have determined that the value of k that makes the statement true is:

k = 1

This solution demonstrates the importance of understanding the rules of exponents and using algebraic techniques to solve equations. By breaking down the problem into smaller, manageable steps, we were able to systematically find the value of the unknown variable. This process not only provides us with the answer but also enhances our problem-solving skills and our appreciation for the elegance of mathematical reasoning. The journey of solving for k has been a testament to the power of algebra and its ability to unravel complex equations.

Conclusion

In conclusion, we have successfully navigated the algebraic landscape and found the value of k in the equation x^k y^4(2x3 + 7x^2y4) = 2x^4y4 + 7x^3y8. Our journey involved applying the distributive property, simplifying expressions with exponents, and comparing terms on both sides of the equation. Through these steps, we discovered that k = 1 is the solution that makes the statement true. This exercise has not only provided us with a specific answer but has also reinforced our understanding of fundamental algebraic principles and problem-solving strategies. The ability to manipulate equations and solve for unknown variables is a crucial skill in mathematics and beyond, and this exploration has served as a valuable learning experience. The world of algebra is filled with intriguing challenges, and by embracing these challenges with a clear understanding of the rules and a methodical approach, we can unlock their secrets and appreciate the beauty of mathematical solutions.