Solving For S In The Exponential Equation (1/64) = 4^(2s-1) * 16^(2s+2)

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In this article, we will delve into the process of solving for the unknown variable s in the exponential equation: $\frac{1}{64}=4^{2 s-1} \cdot 16^{2 s+2}$. This problem is a classic example of how to manipulate exponential expressions using the properties of exponents. Understanding these properties is crucial for solving various mathematical problems, especially in algebra and calculus. We will break down the equation step by step, transforming it into a more manageable form and ultimately isolating s to find its value. This comprehensive guide aims to provide clarity and build confidence in handling exponential equations.

Understanding Exponential Equations

Before diving into the specific problem, let's first establish a solid understanding of exponential equations. Exponential equations are equations where the variable appears in the exponent. The key to solving these equations lies in the properties of exponents, which allow us to manipulate and simplify expressions. One of the most important properties is the ability to express numbers with the same base. When we have exponential terms with the same base, we can equate the exponents if the bases are equal. This principle forms the foundation of our approach to solving the given equation. Another crucial property is the power of a power rule, which states that (am)n=am∗n(a^m)^n = a^{m*n}. This rule helps us simplify complex exponential expressions by multiplying exponents. Furthermore, understanding negative exponents is essential; a negative exponent indicates a reciprocal, such as a−n=1ana^{-n} = \frac{1}{a^n}. By mastering these fundamental properties, we can effectively transform and solve a wide range of exponential equations. In the context of this problem, we will leverage these properties to rewrite the equation in a simplified form, allowing us to isolate the variable s. The process involves expressing all terms with the same base, simplifying exponents, and then solving the resulting algebraic equation. The ultimate goal is to provide a clear and concise method for solving exponential equations, empowering readers to tackle similar problems with ease and confidence.

Step 1: Express all Terms with the Same Base

The first crucial step in solving the equation $\frac{1}{64}=4^{2 s-1} \cdot 16^{2 s+2}$ is to express all terms with the same base. Recognizing that both 64 and 16 are powers of 4 is the key here. We can rewrite 64 as 434^3 and 16 as 424^2. This transformation allows us to bring all parts of the equation into a common exponential form, making it easier to manipulate and simplify. By expressing the numbers with the same base, we can directly compare and equate exponents, which is a fundamental technique in solving exponential equations. Let's start by rewriting the left side of the equation. Since $\frac{1}{64}$ is the reciprocal of 64, we can express it as 64−164^{-1}. Substituting 434^3 for 64, we get (43)−1(4^3)^{-1}, which simplifies to 4−34^{-3} using the power of a power rule. Now, let's focus on the right side of the equation. We have 42s−14^{2s-1} already in base 4, so we leave it as is. For the term 162s+216^{2s+2}, we substitute 424^2 for 16, resulting in (42)2s+2(4^2)^{2s+2}. Applying the power of a power rule again, we multiply the exponents, yielding 42(2s+2)4^{2(2s+2)}, which simplifies to 44s+44^{4s+4}. With both sides of the equation now expressed in base 4, we can rewrite the entire equation as 4−3=42s−1⋅44s+44^{-3} = 4^{2s-1} \cdot 4^{4s+4}. This unified base sets the stage for the next step, where we will combine the exponential terms on the right side of the equation, further simplifying it and bringing us closer to isolating the variable s. The ability to recognize and manipulate bases is a critical skill in solving exponential equations, and this step highlights its importance in simplifying complex problems.

Step 2: Combine Exponential Terms

With the equation now in the form 4−3=42s−1⋅44s+44^{-3} = 4^{2s-1} \cdot 4^{4s+4}, the next step involves combining the exponential terms on the right side. To do this, we utilize the property of exponents that states when multiplying exponential terms with the same base, we add the exponents. This property, expressed as am⋅an=am+na^m \cdot a^n = a^{m+n}, is a fundamental rule in simplifying exponential expressions and is crucial for solving equations of this nature. Applying this rule to our equation, we combine 42s−14^{2s-1} and 44s+44^{4s+4} by adding their exponents: (2s−1)+(4s+4)(2s - 1) + (4s + 4). Simplifying this sum gives us 6s+36s + 3. Therefore, the right side of the equation becomes 46s+34^{6s+3}. Now, the equation is significantly simplified, transforming from its initial complex form to 4−3=46s+34^{-3} = 4^{6s+3}. This simplification is a pivotal moment in the solving process because it allows us to directly compare the exponents on both sides of the equation. By reducing the equation to this simpler form, we eliminate the exponential terms and pave the way for solving a straightforward linear equation. The ability to combine exponential terms effectively is a key skill in handling exponential equations, and this step demonstrates how it can dramatically simplify complex expressions. The next step will involve equating the exponents, which will lead us to a simple algebraic equation that we can easily solve for the variable s. The process of combining exponential terms not only simplifies the equation but also brings us closer to our ultimate goal of isolating and determining the value of s.

Step 3: Equate the Exponents

Having simplified the equation to 4−3=46s+34^{-3} = 4^{6s+3}, the next logical step is to equate the exponents. This is a direct application of the principle that if am=ana^m = a^n, then m=nm = n, provided that the base a is the same on both sides of the equation. In our case, the base is 4, and it is the same on both sides, so we can confidently equate the exponents. By equating the exponents, we transform the exponential equation into a simple linear equation, which is much easier to solve. This is a critical step in the solution process, as it eliminates the exponential terms and allows us to focus on the algebraic manipulation needed to isolate the variable s. Equating the exponents, we get the linear equation −3=6s+3-3 = 6s + 3. This equation represents the core relationship that we need to solve for s. The left side of the equation, -3, is the exponent from the term 4−34^{-3}, and the right side, 6s+36s + 3, is the exponent from the term 46s+34^{6s+3}. By setting these two expressions equal to each other, we create a direct link between the variable s and the constant values in the equation. This step highlights the power of exponential properties in simplifying complex equations. By understanding and applying these properties, we can reduce a seemingly complicated problem to a manageable algebraic form. The next step will involve solving this linear equation for s, which will provide us with the final solution to the original exponential equation. The ability to equate exponents is a fundamental technique in solving exponential equations, and this step illustrates its importance in transforming and simplifying problems.

Step 4: Solve for s

Now that we have the linear equation −3=6s+3-3 = 6s + 3, we can solve for s. This involves isolating s on one side of the equation by performing algebraic manipulations. The first step in solving for s is to subtract 3 from both sides of the equation. This operation maintains the balance of the equation while moving the constant term from the right side to the left side. Subtracting 3 from both sides gives us: -3 - 3 = 6s + 3 - 3, which simplifies to -6 = 6s. Now, we have a simpler equation with the s term isolated on one side. The next step is to divide both sides of the equation by 6. This will isolate s completely, giving us the value of s. Dividing both sides by 6, we get: $ rac{-6}{6} = \frac{6s}{6}$, which simplifies to -1 = s. Therefore, the solution to the equation is s = -1. This is the final value of s that satisfies the original exponential equation. The process of solving this linear equation demonstrates the fundamental principles of algebra, where we use inverse operations to isolate the variable of interest. By subtracting and dividing, we systematically removed the terms surrounding s until we were left with s alone. This step highlights the importance of algebraic skills in solving mathematical problems, particularly in the context of exponential equations. With s now determined, we have successfully solved the original equation. The ability to solve for variables in linear equations is a crucial skill in mathematics, and this step reinforces its application in the context of exponential equations. The final step will be to verify our solution, ensuring that it satisfies the original equation and confirming the accuracy of our calculations.

Step 5: Verify the Solution

To ensure the accuracy of our solution, it is crucial to verify the value of s we found, which is s = -1. Verification involves substituting the value of s back into the original equation and checking if both sides of the equation are equal. This step is a critical safeguard against errors that may have occurred during the solving process. Substituting s = -1 into the original equation $\frac1}{64}=4^{2 s-1} \cdot 16^{2 s+2}$, we get $\frac{164}=4^{2(-1)-1} \cdot 16^{2(-1)+2}$. Now, we simplify the exponents $\frac{164}=4^{-2-1} \cdot 16^{-2+2}$, which further simplifies to $\frac{164}=4^{-3} \cdot 16^{0}$. Recall that any non-zero number raised to the power of 0 is 1, so 160=116^0 = 1. Also, 4−34^{-3} is the reciprocal of 434^3, which is 64, so 4−3=1644^{-3} = \frac{1}{64}. Substituting these values, we get $\frac{164}=\frac{1}{64} \cdot 1$, which simplifies to $\frac{1{64}=\frac{1}{64}$. Since both sides of the equation are equal, our solution s = -1 is verified to be correct. This verification step not only confirms the accuracy of our calculations but also reinforces our understanding of the properties of exponents and how they apply in solving equations. By systematically substituting the solution back into the original equation, we can confidently assert that our answer is correct. Verification is a fundamental aspect of problem-solving in mathematics, and this step highlights its importance in ensuring the validity of our results. In this case, the successful verification of s = -1 completes the solution process, providing a comprehensive and accurate answer to the original problem.

Conclusion

In conclusion, we have successfully solved the exponential equation $\frac{1}{64}=4^{2 s-1} \cdot 16^{2 s+2}$ by systematically applying the properties of exponents and algebraic manipulation techniques. The solution, s = -1, was obtained through a series of steps, each building upon the previous one. First, we expressed all terms with the same base, which allowed us to combine exponential terms and simplify the equation. Then, we equated the exponents, transforming the exponential equation into a linear equation. Solving this linear equation gave us the value of s, which we subsequently verified by substituting it back into the original equation. This step-by-step process demonstrates the power of understanding and applying mathematical principles to solve complex problems. By breaking down the equation into manageable parts and utilizing the properties of exponents, we were able to isolate the variable s and determine its value accurately. This comprehensive approach not only provides the solution but also enhances our understanding of exponential equations and their solutions. The ability to solve exponential equations is a valuable skill in mathematics, with applications in various fields such as physics, engineering, and finance. By mastering the techniques outlined in this article, readers can confidently tackle similar problems and further develop their mathematical problem-solving abilities. The verification step underscores the importance of accuracy and attention to detail in mathematics, ensuring that the solution obtained is indeed correct. The final answer, s = -1, represents the culmination of a methodical and rigorous approach to solving exponential equations.