Finding The Value Of K In The Equation (5a²b³)(6aᵏb) = 30a⁶b⁴
Introduction
In the realm of algebraic equations, solving for unknown variables is a fundamental skill. This article delves into a specific problem where we aim to find the value of k that satisfies the equation (5a²b³)(6aᵏb) = 30a⁶b⁴. This equation involves the multiplication of terms with exponents, making it a great example of applying the rules of exponents. By understanding these rules and carefully manipulating the equation, we can isolate k and determine its value. This exploration is not just an exercise in algebra; it's a journey into the heart of mathematical problem-solving, where logic and precision are key. Mastering such problems builds a solid foundation for tackling more complex mathematical challenges in the future. So, let's embark on this algebraic adventure and uncover the mystery of k.
Understanding the Equation
To effectively solve the equation (5a²b³)(6aᵏb) = 30a⁶b⁴, we first need to understand its structure and the underlying mathematical principles. The equation involves the product of two algebraic expressions on the left-hand side, which equals a single expression on the right-hand side. Each expression consists of coefficients (the numerical parts) and variables (a and b) raised to certain powers (exponents). The key to solving for k lies in applying the rules of exponents, specifically the rule for multiplying terms with the same base. This rule states that when multiplying exponential terms with the same base, we add their exponents. For example, aᵐ * a*ⁿ = a⁽ᵐ⁺ⁿ⁾. Applying this rule, we can simplify the left-hand side of the equation by combining the a terms and the b terms. This process will lead us to an equation where we can equate the exponents of a and b on both sides, allowing us to solve for the unknown k. Understanding these fundamental concepts is crucial for not only solving this particular problem but also for tackling a wide range of algebraic equations involving exponents.
Step-by-Step Solution
Let's embark on a step-by-step journey to unravel the value of k in the equation (5a²b³)(6aᵏb) = 30a⁶b⁴. This process will not only reveal the solution but also illuminate the elegance of algebraic manipulation. Our journey begins with a careful application of the commutative and associative properties of multiplication, allowing us to rearrange and regroup the terms. This initial step is crucial as it sets the stage for simplifying the equation. Next, we harness the power of the exponent rules, specifically the rule that dictates when multiplying terms with the same base, we add their exponents. This rule is the linchpin that connects the left-hand side of the equation to the right-hand side. By meticulously applying this rule, we transform the equation into a more manageable form, where the exponents of a and b on both sides become comparable. The subsequent step involves equating the exponents of the like bases, a critical juncture where we transition from a single equation with multiple terms to a set of simpler equations. These simpler equations are the key to unlocking the value of k. Solving these equations is the final step in our journey, and it is here that we finally unveil the numerical value of k that satisfies the original equation. Each step in this process is a testament to the methodical approach required in algebraic problem-solving, a journey that culminates in a satisfying resolution.
1. Combine the Coefficients
The first step in simplifying the equation (5a²b³)(6aᵏb) = 30a⁶b⁴ is to combine the numerical coefficients. We have 5 and 6 as the coefficients on the left-hand side. Multiplying these together, we get:
5 * 6 = 30
This simplifies our equation to:
30(a²b³)(aᵏb) = 30a⁶b⁴
2. Apply the Product of Powers Rule
Now, let's focus on the variables with exponents. We have a² and aᵏ, as well as b³ and b. According to the product of powers rule, when multiplying terms with the same base, we add their exponents. Applying this rule, we get:
a² * aᵏ = a⁽²⁺ᵏ⁾ b³ * b = b⁽³⁺¹⁾ = b⁴
Substituting these back into the equation, we now have:
30a⁽²⁺ᵏ⁾b⁴ = 30a⁶b⁴
3. Equate the Exponents
With the equation simplified, we can now equate the exponents of the corresponding variables on both sides. We have:
For a: 2 + k = 6 For b: 4 = 4 (This confirms our simplification is correct for b)
4. Solve for k
To find the value of k, we solve the equation:
2 + k = 6
Subtracting 2 from both sides, we get:
k = 6 - 2 k = 4
Therefore, the value of k that makes the equation true is 4.
Verification
To ensure our solution is correct, we must embark on a crucial step: verification. This process is not merely a formality; it's the cornerstone of mathematical rigor, the ultimate test of our algebraic journey. We substitute the value we found for k back into the original equation, transforming the unknown into a known entity. This act of substitution is akin to placing the final piece in a puzzle, completing the picture and revealing whether our solution aligns with the initial conditions. With k replaced by its numerical value, we simplify both sides of the equation, meticulously applying the order of operations and the rules of exponents. This simplification process is a delicate dance of mathematical symbols, each step carefully executed to maintain equality and accuracy. The moment of truth arrives when both sides of the equation converge to the same expression. This convergence is not just a numerical coincidence; it's a resounding confirmation that our solution is indeed correct. It's a testament to the power of algebraic manipulation and the precision of our step-by-step approach. Verification, therefore, is not just a step; it's the seal of approval on our mathematical endeavor, solidifying our understanding and confidence in the solution.
Substituting k = 4 back into the original equation, we have:
(5a²b³)(6a⁴b) = 30a⁶b⁴
Simplifying the left-hand side:
30a⁽²⁺⁴⁾b⁽³⁺¹⁾ = 30a⁶b⁴ 30a⁶b⁴ = 30a⁶b⁴
The left-hand side equals the right-hand side, so our solution k = 4 is correct.
Conclusion
In conclusion, we have successfully navigated the algebraic landscape of the equation (5a²b³)(6aᵏb) = 30a⁶b⁴, and through our step-by-step exploration, we have unveiled the value of k that renders the equation true. This journey has been more than just a mathematical exercise; it has been a testament to the power of algebraic manipulation and the elegance of mathematical reasoning. We began by understanding the equation's structure, dissecting its components, and recognizing the role of coefficients, variables, and exponents. This initial comprehension laid the groundwork for our subsequent steps. We then strategically applied the rules of exponents, specifically the product of powers rule, to simplify the equation. This application was not merely a mechanical process; it was a thoughtful transformation, guiding us towards a more manageable form. The pivotal moment arrived when we equated the exponents of like bases, a step that translated the original equation into a simpler, more accessible form. This simplification was the key that unlocked the value of k. The final act of solving for k was the culmination of our efforts, revealing the numerical answer that satisfied the equation. But our journey did not end there. We embraced the principle of mathematical rigor and subjected our solution to the test of verification. This crucial step confirmed the accuracy of our answer, reinforcing our understanding and solidifying our confidence. The value of k = 4 is not just a number; it's the embodiment of our algebraic exploration, a symbol of our mathematical journey.
