Solving Exponential Equations Finding The Value Of A In (1/9)^(a+1) = 81^(a+1) * 27^(a-3)

Jul 14, 2025 by ADMIN 90 views

Solving for 'a' in the Exponential Equation: A Comprehensive Guide

Introduction

In this article, we will delve into the process of determining the value of 'a' that satisfies the exponential equation: (1/9)^(a+1) = 81^(a+1) * 27^(a-3). This problem falls under the category of exponential equations, which are equations where the variable appears in the exponent. Solving such equations often involves manipulating the expressions to have the same base, and then equating the exponents. We will explore the fundamental concepts of exponents and logarithms, and apply these concepts to systematically solve the given equation. Our discussion will cover the properties of exponents, the process of simplification, and the algebraic manipulations required to isolate the variable 'a'. We will also highlight common pitfalls and strategies to avoid them, ensuring a clear and concise understanding of the solution. This exploration is crucial not only for solving this specific problem but also for building a strong foundation in algebra and exponential functions, which are essential in various fields of mathematics, science, and engineering. Understanding how to solve exponential equations like this one is a fundamental skill in algebra, providing a pathway to more complex mathematical problems and real-world applications.

Understanding the Fundamentals of Exponential Equations

Before we dive into the solution, let's reinforce our understanding of exponential equations and the properties of exponents. An exponential equation is an equation in which the variable occurs in the exponent. The key to solving these equations lies in the ability to manipulate expressions so that they share a common base. When expressions with the same base are equated, we can equate their exponents, thereby transforming the exponential equation into a simpler algebraic equation. Properties of exponents, such as the product of powers, quotient of powers, and power of a power, play a crucial role in this manipulation. For instance, the rule (x^m)n = x^(mn)* allows us to simplify expressions raised to a power. Similarly, the rule x^m * x^n = x^(m+n) helps us combine terms with the same base. We will utilize these properties extensively in the process of solving the equation. A solid grasp of these exponential rules is crucial for navigating the complexities of the equation and arriving at the correct value for 'a'. Recognizing the relationships between different bases, like 9, 81, and 27, as powers of a common base (in this case, 3), is a critical step in simplifying the equation and ultimately solving for 'a'. This foundational understanding allows us to approach the problem methodically and with confidence.

Step-by-Step Solution

Now, let's proceed with the step-by-step solution of the equation (1/9)^(a+1) = 81^(a+1) * 27^(a-3). The first step is to express all terms with the same base. We recognize that 9, 81, and 27 can all be written as powers of 3. Specifically, 9 = 3^2, 81 = 3^4, and 27 = 3^3. Rewriting the equation using base 3, we get (3^(-2))(a+1) = (3⁴⁾(a+1) * (3³⁾(a-3). Next, we apply the power of a power rule, which states that (x^m)n = x^(mn)*. This gives us 3^(-2(a+1)) = 3^(4(a+1)) * 3^(3(a-3)). Now, we simplify the exponents: 3^(-2a-2) = 3^(4a+4) * 3^(3a-9). We then apply the product of powers rule, which states that x^m * x^n = x^(m+n), to the right side of the equation. This results in 3^(-2a-2) = 3^(4a+4+3a-9). Simplifying the exponent on the right side, we have 3^(-2a-2) = 3^(7a-5). Since the bases are now the same, we can equate the exponents: -2a - 2 = 7a - 5. This transforms the exponential equation into a linear equation, which is much easier to solve. Adding 2a to both sides and adding 5 to both sides, we get 3 = 9a. Finally, dividing both sides by 9, we find that a = 3/9 = 1/3. Therefore, the value of a that satisfies the given equation is 1/3. This systematic approach ensures accuracy and clarity in the solution.

Detailed Breakdown of Each Step

To further clarify the solution process, let's provide a detailed breakdown of each step involved in solving the equation (1/9)^(a+1) = 81^(a+1) * 27^(a-3). This breakdown will not only reinforce the concepts used but also help in understanding the logic behind each manipulation. The initial step involves recognizing the common base among the terms. We identified that 9, 81, and 27 can all be expressed as powers of 3. This is a crucial observation that simplifies the equation significantly. By expressing 1/9 as 3^(-2), 81 as 3^4, and 27 as 3^3, we lay the foundation for applying the properties of exponents. The next step is the application of the power of a power rule. This rule allows us to simplify expressions like (3^(-2))(a+1) into 3^(-2(a+1)). Similarly, (3⁴⁾(a+1) becomes 3^(4(a+1)), and (3³⁾(a-3) becomes 3^(3(a-3)). This step reduces the complexity of the exponents and prepares the equation for the next phase. Following this, we simplify the exponents further by distributing the multiplication. For instance, -2(a+1) becomes -2a - 2. This simplification is crucial for combining like terms later in the process. The application of the product of powers rule is another key step. This rule allows us to combine terms with the same base by adding their exponents. So, 3^(4a+4) * 3^(3a-9) becomes 3^(4a+4+3a-9). This step consolidates the exponential terms on one side of the equation. Once the exponential terms are combined, we simplify the exponent by adding like terms. In this case, 4a + 4 + 3a - 9 simplifies to 7a - 5. This simplification results in an equation of the form 3^(-2a-2) = 3^(7a-5). The pivotal step of equating the exponents comes next. Since the bases are the same, we can equate the exponents, transforming the exponential equation into a linear equation: -2a - 2 = 7a - 5. This step significantly simplifies the problem. The final steps involve solving the linear equation for 'a'. This is done by isolating 'a' on one side of the equation. By adding 2a to both sides and adding 5 to both sides, we get 3 = 9a. Then, dividing both sides by 9, we find that a = 1/3. This detailed breakdown not only provides a step-by-step guide but also emphasizes the underlying principles and properties of exponents that are essential for solving exponential equations.

Common Mistakes and How to Avoid Them

Solving exponential equations can be tricky, and there are several common mistakes that students often make. Understanding these pitfalls and learning how to avoid them is crucial for achieving accurate solutions. One common mistake is incorrectly applying the properties of exponents. For instance, students may mistakenly add exponents when the bases are different, or they might forget to distribute exponents when dealing with powers of powers. To avoid this, it's essential to have a solid understanding of the exponent rules and to practice applying them in various contexts. Another frequent error is failing to express all terms with a common base before equating exponents. This can lead to incorrect simplifications and ultimately, the wrong answer. Always ensure that all terms are expressed with the same base before proceeding with the equation. A third mistake is making errors in algebraic manipulation, particularly when solving the linear equation that results after equating exponents. This can include mistakes in adding, subtracting, multiplying, or dividing terms. To minimize these errors, it's crucial to double-check each step of the algebraic process and to work neatly and systematically. Another pitfall is forgetting to account for negative exponents. Negative exponents indicate reciprocals, and it's important to handle them correctly. For example, 1/9 should be recognized as 3^(-2), not just 3^(-1). Finally, a common mistake is not checking the solution in the original equation. Plugging the value of 'a' back into the original equation can help identify any errors made during the solving process. By being aware of these common mistakes and taking steps to avoid them, students can significantly improve their accuracy and confidence in solving exponential equations.

Practical Applications of Exponential Equations

Exponential equations are not just abstract mathematical concepts; they have numerous practical applications in various fields, including science, engineering, finance, and computer science. Understanding how to solve them is therefore essential for real-world problem-solving. In science, exponential equations are used to model phenomena such as population growth, radioactive decay, and chemical reactions. For example, the decay of a radioactive substance can be modeled using an exponential equation, where the amount of the substance decreases exponentially over time. Similarly, the growth of a population can be modeled using an exponential equation, where the population increases exponentially under certain conditions. In finance, exponential equations are used to calculate compound interest, which is the interest earned on both the principal amount and the accumulated interest. Understanding compound interest is crucial for making informed financial decisions, such as investments and loans. In engineering, exponential equations are used in various applications, including circuit analysis and signal processing. For instance, the charging and discharging of a capacitor in an electrical circuit can be modeled using an exponential equation. In computer science, exponential equations are used in the analysis of algorithms and data structures. The time complexity of certain algorithms can be expressed using exponential functions, which helps in understanding the efficiency of the algorithm. Furthermore, exponential functions are used in cryptography, where they play a role in encoding and decoding messages. The ubiquity of exponential equations in these diverse fields underscores the importance of mastering their solution techniques. By understanding the practical applications of these equations, students can appreciate the relevance of the mathematical concepts they are learning and be better prepared for future challenges in their respective fields. The ability to solve exponential equations is a valuable skill that extends far beyond the classroom.

Conclusion

In conclusion, solving the exponential equation (1/9)^(a+1) = 81^(a+1) * 27^(a-3) involves a systematic application of the properties of exponents and algebraic manipulation. By expressing all terms with a common base, applying the power of a power rule, simplifying exponents, and equating exponents, we transformed the exponential equation into a linear equation. Solving this linear equation, we found that a = 1/3. This process highlights the importance of a strong foundation in exponent rules and algebraic techniques. Furthermore, we discussed common mistakes that students often make when solving exponential equations, such as incorrectly applying exponent rules or failing to express terms with a common base. By being aware of these pitfalls and practicing strategies to avoid them, students can improve their accuracy and confidence in solving these types of problems. We also explored the practical applications of exponential equations in various fields, including science, engineering, finance, and computer science. These applications underscore the relevance of exponential equations in real-world problem-solving. The ability to solve exponential equations is a valuable skill that extends far beyond the classroom. Mastering this skill not only enhances mathematical proficiency but also prepares individuals for future challenges in their respective fields. By understanding the concepts and techniques discussed in this article, readers can confidently approach and solve a wide range of exponential equations.