Solving (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ Exponential Equation

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In this article, we will delve into solving the exponential equation (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ. This type of problem often appears in mathematics, specifically within the realms of algebra and exponents. Understanding how to manipulate exponents and fractions is crucial for tackling such equations. We will break down the problem step by step, explaining the underlying principles and mathematical operations involved. Our goal is not just to find the value of 'x' but also to provide a comprehensive understanding of the methods used, which can be applied to similar problems. By the end of this discussion, you should be well-equipped to solve similar exponential equations and have a solid grasp of the fundamental concepts involved. This includes the rules of exponents, fraction manipulation, and how to equate exponential expressions with the same base. Let's embark on this mathematical journey and unlock the solution together.

Understanding the Basics of Exponents

Before we dive into the solution, it's essential to have a firm grasp of the basic rules of exponents. Exponents, also known as powers, represent the number of times a base number is multiplied by itself. For example, in the expression aⁿ, 'a' is the base, and 'n' is the exponent. This means 'a' is multiplied by itself 'n' times. Some key rules of exponents that will be crucial for solving our equation include:

  1. Product of Powers: aᵐ × aⁿ = aᵐ⁺ⁿ (When multiplying powers with the same base, add the exponents).
  2. Quotient of Powers: aᵐ / aⁿ = aᵐ⁻ⁿ (When dividing powers with the same base, subtract the exponents).
  3. Power of a Power: (aᵐ)ⁿ = aᵐⁿ (When raising a power to another power, multiply the exponents).
  4. Negative Exponent: a⁻ⁿ = 1/aⁿ (A negative exponent indicates the reciprocal of the base raised to the positive exponent).
  5. Zero Exponent: a⁰ = 1 (Any non-zero number raised to the power of 0 is 1).

These rules form the foundation for manipulating exponential expressions and solving exponential equations. We will apply these rules strategically to simplify the given equation and isolate the variable 'x'. Remember, the key to solving exponential equations is often to express both sides of the equation with the same base. This allows us to equate the exponents and solve for the unknown. With these fundamentals in mind, let's proceed to the step-by-step solution of our equation.

Step-by-Step Solution

Let's start by rewriting the given equation: (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ. The first step involves simplifying the left-hand side (LHS) of the equation. Notice that we have (3/4)³ and (4/3)⁻⁷. To effectively combine these terms, we need to express them with the same base. Since 4/3 is the reciprocal of 3/4, we can use the rule of negative exponents to rewrite (4/3)⁻⁷. Recall that a⁻ⁿ = 1/aⁿ, so (4/3)⁻⁷ can be written as (3/4)⁷. Now, our equation looks like this: (3/4)³ × (3/4)⁷ = (9/16)ˣ. We can now apply the product of powers rule, which states that aᵐ × aⁿ = aᵐ⁺ⁿ. In our case, the base is 3/4, and the exponents are 3 and 7. So, we add the exponents: (3/4)³⁺⁷ = (3/4)¹⁰. Now our equation is: (3/4)¹⁰ = (9/16)ˣ. The next step is to express the right-hand side (RHS) of the equation, (9/16)ˣ, in terms of the base 3/4. We know that 9 is 3² and 16 is 4². Therefore, 9/16 can be written as (3²/4²), which is the same as (3/4)². So, (9/16)ˣ can be rewritten as ((3/4)²)ˣ. Now we apply the power of a power rule, (aᵐ)ⁿ = aᵐⁿ, to simplify ((3/4)²)ˣ, which becomes (3/4)²ˣ. Our equation now reads: (3/4)¹⁰ = (3/4)²ˣ. We have successfully expressed both sides of the equation with the same base (3/4). Now we can equate the exponents. This means that the exponent on the LHS must be equal to the exponent on the RHS. Therefore, 10 = 2x. To solve for 'x', we divide both sides of the equation by 2: x = 10/2, which simplifies to x = 5. Therefore, the solution to the equation (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ is x = 5.

Verification of the Solution

To ensure our solution is correct, it's crucial to verify it by substituting the value of 'x' back into the original equation. Our solution is x = 5, and the original equation is (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ. Substituting x = 5 into the equation, we get: (3/4)³ × (4/3)⁻⁷ = (9/16)⁵. We've already simplified the left-hand side (LHS) of the equation in our earlier steps, and we know that (3/4)³ × (4/3)⁻⁷ simplifies to (3/4)¹⁰. Now, let's simplify the right-hand side (RHS) of the equation, (9/16)⁵. We know that 9/16 is (3/4)², so (9/16)⁵ can be written as ((3/4)²)⁵. Applying the power of a power rule, (aᵐ)ⁿ = aᵐⁿ, we get (3/4)²*⁵ = (3/4)¹⁰. Now, we compare the LHS and RHS: LHS = (3/4)¹⁰ RHS = (3/4)¹⁰. Since both sides are equal, our solution x = 5 is verified. This verification step is essential in mathematics to confirm that the calculated solution satisfies the original equation. It helps to eliminate errors and ensures the accuracy of the answer. In this case, our verification confirms that x = 5 is indeed the correct solution to the exponential equation.

Alternative Approaches to Solving Exponential Equations

While we have successfully solved the equation (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ using a step-by-step simplification approach, it's worth exploring alternative methods that can be applied to solve similar exponential equations. One such method involves the use of logarithms. Logarithms provide a powerful tool for solving equations where the variable appears in the exponent. The basic principle behind using logarithms is to transform an exponential equation into a linear equation, which is generally easier to solve. To illustrate, let's consider our original equation: (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ. After simplifying the LHS as before, we have (3/4)¹⁰ = (9/16)ˣ. We know that 9/16 can be written as (3/4)², so the equation becomes (3/4)¹⁰ = ((3/4)²)ˣ, which simplifies to (3/4)¹⁰ = (3/4)²ˣ. At this point, we equated the exponents directly because the bases were the same. However, let's see how logarithms could be applied. If we take the logarithm of both sides of the equation (3/4)¹⁰ = (9/16)ˣ, we can use the logarithmic property log(aᵇ) = b * log(a). Applying this property, we get: 10 * log(3/4) = x * log(9/16). Now, we can solve for 'x' by dividing both sides by log(9/16): x = (10 * log(3/4)) / log(9/16). Using a calculator, we can evaluate the logarithms and find the value of 'x'. Another approach involves converting the equation to a common base from the beginning. In our case, we could express all terms in terms of powers of 2 and 3, which might lead to a different simplification path but ultimately the same solution. Exploring these alternative methods not only provides different perspectives on solving exponential equations but also enhances problem-solving skills and mathematical flexibility. Each method may have its advantages depending on the specific equation, and understanding multiple approaches can be beneficial in tackling a variety of mathematical problems.

Common Mistakes to Avoid

When solving exponential equations, several common mistakes can lead to incorrect solutions. Being aware of these pitfalls can help you avoid them and improve your accuracy. One frequent error is misapplying the rules of exponents. For example, students sometimes incorrectly add exponents when the bases are different, or they might try to multiply exponents when they should be adding them. Remember, the rule aᵐ × aⁿ = aᵐ⁺ⁿ only applies when the bases are the same. Another common mistake is mishandling negative exponents. A negative exponent indicates the reciprocal of the base raised to the positive exponent, i.e., a⁻ⁿ = 1/aⁿ. Forgetting this rule or applying it incorrectly can lead to significant errors. Similarly, the zero exponent rule, a⁰ = 1, is often overlooked or misapplied. Another area where mistakes frequently occur is in simplifying fractions. When dealing with fractional bases and exponents, it's crucial to simplify fractions correctly and apply exponent rules to both the numerator and the denominator. For instance, (a/b)ⁿ = aⁿ/bⁿ. A significant mistake that can arise is not verifying the solution. After solving for the variable, it's essential to substitute the value back into the original equation to ensure it holds true. This step can catch algebraic errors or incorrect simplifications made during the solving process. In the context of using logarithms, a common mistake is misapplying logarithmic properties or using the calculator incorrectly to evaluate logarithms. It's important to understand the properties of logarithms and use them correctly to transform the equation and solve for the variable. Additionally, when taking logarithms, ensure you use the same base for all terms in the equation. Finally, rushing through the steps or not paying close attention to detail can lead to simple arithmetic errors. Taking your time, writing out each step clearly, and double-checking your work can significantly reduce the likelihood of making mistakes. By being mindful of these common mistakes and practicing regularly, you can improve your accuracy and confidence in solving exponential equations.

Conclusion

In conclusion, solving the exponential equation (3/4)³ × (4/3)⁻⁷ = (9/16)ˣ involved a series of steps that highlight the importance of understanding and applying the fundamental rules of exponents. We began by simplifying the equation using the properties of exponents, such as the negative exponent rule and the product of powers rule. By expressing both sides of the equation with the same base, we were able to equate the exponents and solve for 'x'. Our step-by-step solution led us to the value of x = 5, which we then verified by substituting it back into the original equation. This verification process is a crucial step in problem-solving as it ensures the accuracy of the solution. We also explored alternative approaches, such as using logarithms, which offer different perspectives and methods for solving exponential equations. Understanding these alternative methods can broaden your mathematical toolkit and enhance your problem-solving skills. Furthermore, we discussed common mistakes to avoid, emphasizing the importance of careful simplification, correct application of exponent rules, and thorough verification of the solution. By being aware of these potential pitfalls, you can minimize errors and improve your proficiency in solving exponential equations. Mastering the techniques and principles discussed in this article not only helps in solving specific equations but also builds a strong foundation for more advanced mathematical concepts. Exponential equations are a fundamental topic in algebra and calculus, and a solid understanding of them is essential for success in higher-level mathematics. Practice is key to mastering these concepts, so working through a variety of problems will further solidify your understanding and skills. With consistent effort and a clear understanding of the underlying principles, you can confidently tackle exponential equations and other mathematical challenges.