Solving Exponential Equations A Step-by-Step Guide To $2^{5x-6} = 1/4$

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Introduction

In this article, we will delve into the process of solving an elementary exponential equation. Exponential equations are a fundamental topic in mathematics, appearing in various fields such as calculus, physics, and engineering. The equation we aim to solve is 25xβˆ’6=142^{5x-6} = \frac{1}{4}. This equation involves an exponential term with a base of 2 and a variable xx in the exponent. Our primary goal is to find the value of xx that satisfies this equation. To achieve this, we will employ several key concepts, including the properties of exponents and the technique of expressing both sides of the equation with the same base. Understanding how to manipulate and solve exponential equations is crucial for anyone studying mathematics or related disciplines. This article provides a step-by-step guide to solving the given equation, making it an invaluable resource for students and enthusiasts alike.

The journey of solving exponential equations begins with a clear understanding of the fundamental principles that govern exponents. Exponents represent the number of times a base is multiplied by itself. For instance, in the term 232^3, the base is 2, and the exponent is 3, meaning 2 is multiplied by itself three times (2 Γ— 2 Γ— 2 = 8). The properties of exponents dictate how these expressions can be manipulated, simplified, and ultimately, how equations involving them can be solved. A key property that we will leverage in solving our equation is the rule that states aβˆ’n=1ana^{-n} = \frac{1}{a^n}. This rule allows us to convert fractions into exponential terms with negative exponents, a crucial step in making both sides of our equation have the same base. Another important concept is the power rule, which states that if am=ana^m = a^n, then m=nm = n. This rule allows us to equate the exponents once we have expressed both sides of the equation with the same base, simplifying the equation into a more manageable linear form.

Before diving into the specific steps, it is essential to emphasize the importance of accuracy and precision in mathematical problem-solving. Each step must be performed carefully, with attention to detail, to ensure the final answer is correct. A small error in the manipulation of exponents or algebraic simplification can lead to a completely different result. Therefore, we will meticulously outline each step, providing clear explanations and justifications along the way. By following this structured approach, readers will not only learn how to solve the equation at hand but also gain a deeper appreciation for the rigor and elegance of mathematical reasoning. In the following sections, we will break down the solution process into manageable parts, starting with rewriting the equation to have a common base, then equating the exponents, and finally solving for xx. Let's embark on this mathematical journey with confidence and precision.

Step-by-Step Solution

Step 1: Rewrite the Equation

The first crucial step in solving the exponential equation 25xβˆ’6=142^{5x-6} = \frac{1}{4} is to rewrite both sides of the equation with a common base. In this case, the base we aim for is 2, as the left side of the equation already has a base of 2. To achieve this on the right side, we need to express 14\frac{1}{4} as a power of 2. Recognizing that 4 is 222^2, we can rewrite 14\frac{1}{4} as 122\frac{1}{2^2}. Now, we can use the property of exponents that states aβˆ’n=1ana^{-n} = \frac{1}{a^n} to further rewrite 122\frac{1}{2^2} as 2βˆ’22^{-2}. This transformation is pivotal because it allows us to express both sides of the equation with the same base, which is essential for solving exponential equations. The rewritten equation now looks like this: 25xβˆ’6=2βˆ’22^{5x-6} = 2^{-2}. This step not only simplifies the equation but also sets the stage for the next step, where we will equate the exponents.

The significance of rewriting the equation with a common base cannot be overstated. Expressing both sides of an exponential equation with the same base is a cornerstone technique in solving such equations. It allows us to directly compare the exponents and form a simpler algebraic equation. Without this step, solving the equation would be significantly more challenging, if not impossible. The ability to recognize and apply this technique is a hallmark of mathematical proficiency. In our case, by transforming 14\frac{1}{4} into 2βˆ’22^{-2}, we have effectively bridged the gap between the two sides of the equation, making them directly comparable. This step demonstrates the power of mathematical manipulation and the elegance of using established rules and properties to simplify complex problems. Furthermore, this initial transformation highlights the importance of having a strong foundation in the properties of exponents. Without a clear understanding of these properties, one might struggle to see the pathway to a solution.

In summary, the first step in solving the exponential equation 25xβˆ’6=142^{5x-6} = \frac{1}{4} involves rewriting the equation so that both sides have a common base. This is achieved by expressing 14\frac{1}{4} as 2βˆ’22^{-2}, resulting in the rewritten equation 25xβˆ’6=2βˆ’22^{5x-6} = 2^{-2}. This step is a critical foundation for the subsequent steps, as it allows us to equate the exponents and solve for the unknown variable xx. The transformation underscores the importance of understanding and applying the properties of exponents in simplifying and solving mathematical problems. With the equation now in this simplified form, we are well-positioned to proceed to the next step, where we will equate the exponents and further simplify the equation.

Step 2: Equate the Exponents

Having successfully rewritten the equation 25xβˆ’6=142^{5x-6} = \frac{1}{4} as 25xβˆ’6=2βˆ’22^{5x-6} = 2^{-2}, the next logical step is to equate the exponents. This step is based on the fundamental property of exponential functions, which states that if am=ana^m = a^n, then m=nm = n, provided that aa is a positive number not equal to 1. In our case, the base aa is 2, which satisfies this condition. Therefore, we can equate the exponents 5xβˆ’65x - 6 and βˆ’2-2, leading to the linear equation 5xβˆ’6=βˆ’25x - 6 = -2. This transformation is a critical juncture in the solution process, as it converts the exponential equation into a simpler, more manageable linear equation. By equating the exponents, we have effectively eliminated the exponential terms, allowing us to focus solely on solving for the variable xx.

The significance of equating exponents lies in its ability to simplify the problem. Exponential equations can often seem daunting due to the presence of variables in the exponents. However, by leveraging the properties of exponents and the underlying nature of exponential functions, we can transform these equations into familiar linear forms. The step of equating exponents is a powerful demonstration of this principle. It showcases how a seemingly complex problem can be reduced to a more elementary one through strategic application of mathematical rules. Moreover, this step highlights the interconnectedness of different mathematical concepts. Exponential equations are not isolated entities; they can be linked to linear equations through the process of equating exponents.

The equation 5xβˆ’6=βˆ’25x - 6 = -2 is now a standard linear equation, which can be solved using basic algebraic techniques. This equation represents a direct relationship between xx and a constant, making it straightforward to isolate xx. The transformation from an exponential equation to a linear equation underscores the power of mathematical abstraction and the ability to represent complex relationships in simpler terms. This step not only moves us closer to the solution but also reinforces the importance of mastering fundamental algebraic skills. In the subsequent step, we will proceed to solve this linear equation, employing the principles of algebraic manipulation to find the value of xx. The process of equating exponents is a pivotal step in our solution, bridging the gap between exponential and linear forms and paving the way for a straightforward solution.

Step 3: Solve for x

After equating the exponents in the equation 25xβˆ’6=2βˆ’22^{5x-6} = 2^{-2}, we arrived at the linear equation 5xβˆ’6=βˆ’25x - 6 = -2. The final step in solving for xx involves isolating xx on one side of the equation. This is a standard procedure in algebra and requires the application of basic algebraic principles. Our goal is to manipulate the equation in such a way that xx is the only term remaining on one side. To begin, we add 6 to both sides of the equation. This operation maintains the equality of the equation while moving the constant term from the left side to the right side. Adding 6 to both sides of 5xβˆ’6=βˆ’25x - 6 = -2 gives us 5xβˆ’6+6=βˆ’2+65x - 6 + 6 = -2 + 6, which simplifies to 5x=45x = 4. Now, we have an equation where xx is multiplied by 5. To isolate xx, we need to divide both sides of the equation by 5. This is the final algebraic manipulation required to solve for xx.

The process of isolating xx in a linear equation is a fundamental skill in algebra. Algebraic manipulation is at the heart of solving equations, and the ability to perform these manipulations accurately and efficiently is crucial. Adding or subtracting the same value from both sides of an equation, or multiplying or dividing both sides by the same non-zero value, are key techniques in this process. In our case, adding 6 to both sides and then dividing by 5 were the specific steps required to isolate xx. Each step is based on the principle of maintaining the balance of the equation, ensuring that the equality remains valid throughout the solution process. The equation 5x=45x = 4 represents a simplified version of the original problem, where xx is directly proportional to a constant. Solving this equation requires only one more step.

Dividing both sides of the equation 5x=45x = 4 by 5 gives us x=45x = \frac{4}{5}. This is the solution to the linear equation and, consequently, the solution to the original exponential equation. The value of xx that satisfies the equation 25xβˆ’6=142^{5x-6} = \frac{1}{4} is 45\frac{4}{5}. This result demonstrates the culmination of all the previous steps, from rewriting the equation with a common base to equating the exponents and finally isolating xx. The solution x=45x = \frac{4}{5} is a precise value, and it can be verified by substituting it back into the original equation to ensure that the equality holds. The successful isolation of xx marks the completion of our solution process. In summary, solving for xx involved adding 6 to both sides of the equation 5xβˆ’6=βˆ’25x - 6 = -2 to get 5x=45x = 4, and then dividing both sides by 5 to obtain x=45x = \frac{4}{5}. This value of xx is the solution to the given exponential equation.

Verification

To ensure the accuracy of our solution, it is crucial to verify that the value we obtained for xx indeed satisfies the original equation. This step provides a check against any potential errors made during the solution process. Our original equation was 25xβˆ’6=142^{5x-6} = \frac{1}{4}, and we found that x=45x = \frac{4}{5}. To verify this solution, we will substitute x=45x = \frac{4}{5} back into the original equation and see if the left-hand side equals the right-hand side. Substituting x=45x = \frac{4}{5} into the left-hand side of the equation gives us 25(45)βˆ’62^{5(\frac{4}{5})-6}. Now, we simplify the exponent. The term 5(45)5(\frac{4}{5}) simplifies to 4, so the exponent becomes 4βˆ’64 - 6, which equals -2. Therefore, the left-hand side of the equation becomes 2βˆ’22^{-2}.

The process of verification is an essential component of mathematical problem-solving. It reinforces the validity of the solution and provides confidence in the accuracy of the calculations. Substituting the obtained value back into the original equation is a direct way to confirm that the solution satisfies the initial conditions of the problem. In our case, by substituting x=45x = \frac{4}{5} into 25xβˆ’6=142^{5x-6} = \frac{1}{4}, we are essentially retracing our steps to ensure that the equality holds. This step not only validates the solution but also deepens our understanding of the problem and the solution process. Verification is not merely a formality; it is an integral part of the problem-solving process that helps to catch errors and build mathematical intuition.

Now, let's continue with our verification. We have simplified the left-hand side of the equation to 2βˆ’22^{-2}. Recall that aβˆ’n=1ana^{-n} = \frac{1}{a^n}, so 2βˆ’22^{-2} is equivalent to 122\frac{1}{2^2}. Since 22=42^2 = 4, we have 2βˆ’2=142^{-2} = \frac{1}{4}. This is exactly the right-hand side of our original equation. Thus, we have shown that substituting x=45x = \frac{4}{5} into the original equation results in a true statement: 25(45)βˆ’6=142^{5(\frac{4}{5})-6} = \frac{1}{4}. This confirms that our solution is correct. The successful verification of the solution underscores the importance of accuracy in each step of the solution process, from rewriting the equation to equating the exponents and solving for xx. In summary, the verification step involved substituting x=45x = \frac{4}{5} back into the original equation, simplifying both sides, and confirming that the equality holds, thereby validating our solution.

Conclusion

In this article, we successfully solved the elementary exponential equation 25xβˆ’6=142^{5x-6} = \frac{1}{4}. The process involved several key steps, starting with rewriting the equation to have a common base, then equating the exponents, and finally solving for xx. We began by recognizing that 14\frac{1}{4} could be expressed as 2βˆ’22^{-2}, which allowed us to rewrite the equation as 25xβˆ’6=2βˆ’22^{5x-6} = 2^{-2}. This step was crucial as it enabled us to equate the exponents, transforming the exponential equation into a linear equation. We then equated the exponents 5xβˆ’65x - 6 and βˆ’2-2, resulting in the linear equation 5xβˆ’6=βˆ’25x - 6 = -2.

Solving for xx involved adding 6 to both sides of the equation, which gave us 5x=45x = 4, and then dividing both sides by 5, yielding the solution x=45x = \frac{4}{5}. This value of xx represents the solution to the original exponential equation. To ensure the accuracy of our solution, we performed a verification step. This involved substituting x=45x = \frac{4}{5} back into the original equation and confirming that the equality holds. The verification process is an essential aspect of mathematical problem-solving, as it helps to identify and correct any potential errors made during the solution process. Our verification confirmed that x=45x = \frac{4}{5} is indeed the correct solution to the equation.

Throughout this article, we have emphasized the importance of understanding and applying the properties of exponents, as well as the techniques for solving linear equations. The solution process highlights the interconnectedness of different mathematical concepts and the importance of a strong foundation in algebra. The ability to rewrite equations, equate exponents, and solve for variables are fundamental skills in mathematics and are essential for success in more advanced topics. Furthermore, the systematic approach we have takenβ€”rewriting, equating, solving, and verifyingβ€”is a valuable problem-solving strategy that can be applied to a wide range of mathematical problems. In conclusion, we have not only solved the given exponential equation but also reinforced the underlying mathematical principles and problem-solving techniques that are crucial for mathematical proficiency.