Solving (8x - 8)^(3/2) = 64 A Step-by-Step Guide

by Aria Freeman 49 views

Hey guys! Today, we're diving into a fun little math problem. We're going to figure out how to solve the equation (8xβˆ’8)32=64(8x - 8)^{\frac{3}{2}} = 64. Sounds a bit intimidating, right? But trust me, we'll break it down step by step, and you'll see it's not so scary after all. We will explore different methods and insights to make sure you understand how to tackle these types of problems. This isn't just about getting the answer; it's about understanding the process. So, grab your pencils, and let's get started!

Understanding the Equation

Before we jump into solving, let’s make sure we really understand what this equation is telling us. The equation (8xβˆ’8)32=64(8x - 8)^{\frac{3}{2}} = 64 might look a bit complex at first glance, but let's break it down. The left side of the equation has a term raised to a fractional exponent, specifically 32\frac{3}{2}. Remember, fractional exponents are just another way of writing radicals. So, (8xβˆ’8)32(8x - 8)^{\frac{3}{2}} can also be written as (8xβˆ’8)3\sqrt{(8x - 8)^3} or (8xβˆ’8)3(\sqrt{8x - 8})^3. Both of these are saying the same thing. Understanding this equivalence is crucial for solving the equation. The right side of the equation is simply 64, a constant value. Our goal is to find the value(s) of 'x' that make the left side equal to 64. This involves a series of algebraic manipulations, which we'll go through methodically. Think of it like peeling an onion – we'll remove each layer step-by-step until we get to the core of the solution. We need to isolate 'x,' and we'll do that by reversing the operations that are being applied to it. So, are you ready to peel this mathematical onion with me? Let's dive into the first step of our solution!

Step-by-Step Solution

Okay, let's get down to business and solve this equation step-by-step. Our initial equation is (8xβˆ’8)32=64(8x - 8)^{\frac{3}{2}} = 64. The first thing we need to tackle is that fractional exponent. Remember, 32\frac{3}{2} means we're dealing with both a power and a root. To get rid of this, we're going to raise both sides of the equation to the power of 23\frac{2}{3}. Why 23\frac{2}{3}? Because when you raise a power to another power, you multiply the exponents. So, (8xβˆ’8)32(8x - 8)^{\frac{3}{2}} raised to the power of 23\frac{2}{3} becomes (8xβˆ’8)(32β‹…23)=(8xβˆ’8)1=8xβˆ’8(8x - 8)^{(\frac{3}{2} \cdot \frac{2}{3})} = (8x - 8)^1 = 8x - 8. This neatly gets rid of the fractional exponent on the left side. On the right side, we have 642364^{\frac{2}{3}}. This might seem tricky, but let's think of it as (643)2(\sqrt[3]{64})^2. The cube root of 64 is 4, and 4 squared is 16. So, our equation now looks like this: 8xβˆ’8=168x - 8 = 16. See? We've already simplified things quite a bit! Now, we just need to isolate 'x'. First, let's add 8 to both sides of the equation: 8xβˆ’8+8=16+88x - 8 + 8 = 16 + 8, which simplifies to 8x=248x = 24. Finally, to solve for 'x', we divide both sides by 8: 8x8=248\frac{8x}{8} = \frac{24}{8}, giving us x=3x = 3. And there you have it! We've found our solution. But hold on, we're not done yet. We always need to verify our solution to make sure it's valid.

Verifying the Solution

Alright, we've got a potential solution: x=3x = 3. But before we declare victory, it's super important to verify that this solution actually works in our original equation. Think of it like this: we've navigated a maze, and now we need to double-check that we've reached the correct exit. Plugging our solution back into the original equation helps us catch any mistakes we might have made along the way, or identify any extraneous solutions (solutions that don't actually satisfy the original equation). So, let's take x=3x = 3 and substitute it into our equation: (8xβˆ’8)32=64(8x - 8)^{\frac{3}{2}} = 64. Replacing 'x' with 3, we get (8(3)βˆ’8)32=64(8(3) - 8)^{\frac{3}{2}} = 64. Let's simplify step-by-step. First, 8(3)=248(3) = 24, so we have (24βˆ’8)32=64(24 - 8)^{\frac{3}{2}} = 64. Next, 24βˆ’8=1624 - 8 = 16, giving us (16)32=64(16)^{\frac{3}{2}} = 64. Now, let's rewrite the fractional exponent as a radical: (16)32(16)^{\frac{3}{2}} is the same as (16)3(\sqrt{16})^3. The square root of 16 is 4, so we have 43=644^3 = 64. And guess what? 434^3 is indeed 64! So, our solution checks out. This verification step is not just a formality; it's a critical part of the problem-solving process. It gives us confidence that our answer is correct and that we haven't made any errors in our calculations or assumptions. Always verify your solutions, guys!

Alternative Approaches

Okay, so we've solved the equation (8xβˆ’8)32=64(8x - 8)^{\frac{3}{2}} = 64 using a straightforward, step-by-step method. But in the world of math, there's often more than one way to skin a cat (or solve an equation!). Exploring alternative approaches can not only deepen our understanding but also give us more tools in our mathematical toolkit. One alternative approach involves rewriting the fractional exponent in its radical form right from the start. Remember, (8xβˆ’8)32(8x - 8)^{\frac{3}{2}} is the same as (8xβˆ’8)3\sqrt{(8x - 8)^3}. So, our equation could be written as (8xβˆ’8)3=64\sqrt{(8x - 8)^3} = 64. From here, we could square both sides of the equation to get rid of the square root: ((8xβˆ’8)3)2=642(\sqrt{(8x - 8)^3})^2 = 64^2, which simplifies to (8xβˆ’8)3=4096(8x - 8)^3 = 4096. Now, we can take the cube root of both sides: (8xβˆ’8)33=40963\sqrt[3]{(8x - 8)^3} = \sqrt[3]{4096}. This gives us 8xβˆ’8=168x - 8 = 16, and from there, we can solve for 'x' as we did before. Another approach might involve factoring out the 8 inside the parentheses initially. We could rewrite the equation as (8(xβˆ’1))32=64(8(x - 1))^{\frac{3}{2}} = 64. Then, using properties of exponents, we can write this as 832(xβˆ’1)32=648^{\frac{3}{2}}(x - 1)^{\frac{3}{2}} = 64. This approach highlights the importance of recognizing and using different mathematical properties and can sometimes lead to a more elegant or efficient solution. Exploring different methods can really help you solidify your understanding and improve your problem-solving skills. So, don't be afraid to try things a different way!

Common Mistakes to Avoid

Now, let's chat about some common pitfalls that people often stumble into when solving equations like this. Knowing these common mistakes can help you steer clear of them and make your problem-solving journey smoother. One frequent error is misinterpreting fractional exponents. Remember, 32\frac{3}{2} means taking the square root and then cubing (or cubing and then taking the square root – the order doesn't matter!). Some folks might get confused and try to multiply the base by the exponent or perform some other incorrect operation. Another mistake is forgetting to apply the exponent to the entire term inside the parentheses. For example, in the expression (8xβˆ’8)32(8x - 8)^{\frac{3}{2}}, you can't simply apply the 32\frac{3}{2} to 8x8x and 8 separately. You have to deal with the entire expression 8xβˆ’88x - 8 as a single unit. Also, a very common oversight is forgetting to verify the solution. We talked about this earlier, but it's worth repeating because it's so crucial. Plugging your solution back into the original equation is your safety net, your way of ensuring that you haven't made a mistake along the way. And finally, be careful with arithmetic errors. Simple mistakes in addition, subtraction, multiplication, or division can throw off your entire solution. Double-checking your calculations is always a good idea. By being aware of these common mistakes, you can approach these types of problems with more confidence and accuracy. So, keep these pitfalls in mind, and you'll be well on your way to becoming a master equation solver!

Conclusion

Alright, guys, we've reached the end of our journey through solving the equation (8xβˆ’8)32=64(8x - 8)^{\frac{3}{2}} = 64. We've broken down the problem step by step, explored alternative approaches, and even discussed common mistakes to avoid. Remember, the key to tackling these types of problems is to understand the underlying concepts, be methodical in your approach, and always verify your solutions. We started by understanding what the equation really meant, particularly those fractional exponents. Then, we systematically isolated 'x' by performing inverse operations. We verified our solution to ensure its validity, and we even looked at different ways we could have approached the problem. This is what mathematical problem-solving is all about – not just getting the answer, but understanding the process. Math isn't about memorizing formulas; it's about developing a way of thinking, a way of approaching challenges. So, keep practicing, keep exploring, and don't be afraid to make mistakes. Mistakes are just opportunities to learn and grow. And most importantly, have fun with it! Math can be a fascinating and rewarding subject, and I hope this guide has helped you feel a little more confident in your equation-solving abilities. Now go out there and conquer those equations!