Solving $(x+15)^{5/3} = 243$: A Step-by-Step Guide

Alex Johnson
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Solving $(x+15)^{5/3} = 243$: A Step-by-Step Guide

Let's dive into solving the equation (x+15)53=243(x+15)^{\frac{5}{3}} = 243. This might seem daunting at first, but by breaking it down step-by-step, we can easily find the value of x. We'll cover everything from understanding fractional exponents to isolating x and verifying our solution. So, grab a pen and paper, and let's get started!

Understanding the Equation

Before we jump into solving, it's crucial to understand what the equation means. The equation (x+15)53=243(x+15)^{\frac{5}{3}} = 243 involves a fractional exponent. Let's break down what a fractional exponent signifies. A fractional exponent like 53\frac{5}{3} represents both a root and a power. The denominator (3 in this case) indicates the type of root we're dealing with, and the numerator (5 in this case) indicates the power to which the base is raised. So, (x+15)53(x+15)^{\frac{5}{3}} can be interpreted as the cube root of (x+15)(x+15) raised to the power of 5, or equivalently, (x+153)5( \sqrt[3]{x+15} )^5. Understanding this is the first key step in solving the equation. The number 243 on the right side is our target value, and our goal is to find the x that makes the left side equal to 243. Recognizing this structure will help us choose the right strategy for solving.

The Role of Fractional Exponents

Fractional exponents are a concise way of expressing both powers and roots. The general form is amna^{\frac{m}{n}}, where a is the base, m is the power, and n is the root. This can be rewritten as (an)m(\sqrt[n]{a})^m or amn\sqrt[n]{a^m}. In our equation, (x+15)53(x+15)^{\frac{5}{3}}, the base is (x+15)(x+15), the power is 5, and the root is 3. Therefore, we are looking for a number which, when 15 is added to it, then we take the cube root, and raise that result to the 5th power, we get 243. Recognizing this relationship between powers and roots is fundamental to solving equations with fractional exponents. This understanding not only simplifies the problem but also guides our steps in isolating x.

Why 243 Matters

The number 243 is significant because it's a power of 3. Specifically, 243=35243 = 3^5. Recognizing this is a critical insight that will simplify our calculations. When we're dealing with fractional exponents, identifying perfect powers on both sides of the equation can help us eliminate the fractional exponent more easily. In this case, knowing that 243 is 353^5 allows us to rewrite the equation in a more manageable form. This connection between 243 and 353^5 allows us to think about the problem in terms of powers and roots of 3, making the solution process clearer and more direct. This step showcases the importance of pattern recognition in solving mathematical problems, a skill that is invaluable in more complex equations as well.

Step-by-Step Solution

Now that we understand the equation, let's proceed with solving it step-by-step:

Step 1: Isolate the Term with the Fractional Exponent

In our equation, (x+15)53=243(x+15)^{\frac{5}{3}} = 243, the term with the fractional exponent is already isolated on the left side. This is a good starting point, as we don't need to perform any additional algebraic manipulations to isolate it. If we had any other terms added or multiplied on the left side, we would need to address those first. But in this case, we can move straight to the next step. Having the term isolated allows us to focus directly on dealing with the fractional exponent, making the process more straightforward. This step highlights the importance of organization in problem-solving, ensuring that the equation is in the most convenient form before proceeding.

Step 2: Raise Both Sides to the Reciprocal Power

To eliminate the fractional exponent 53\frac{5}{3}, we raise both sides of the equation to its reciprocal, which is 35\frac{3}{5}. This is because (amn)nm=a(a^{\frac{m}{n}})^{\frac{n}{m}} = a. Applying this to our equation, we get:

((x+15)53)35=24335((x+15)^{\frac{5}{3}})^{\frac{3}{5}} = 243^{\frac{3}{5}}

On the left side, the exponents 53\frac{5}{3} and 35\frac{3}{5} cancel each other out, leaving us with (x+15)(x+15). On the right side, we have 24335243^{\frac{3}{5}}. Recall that 243=35243 = 3^5, so we can rewrite the right side as (35)35(3^5)^{\frac{3}{5}}. The exponents 5 and 35\frac{3}{5} now interact, simplifying the expression further. This is a crucial step because it eliminates the fractional exponent, bringing us closer to isolating x. Understanding how reciprocal powers work is essential for dealing with fractional exponents, making this step a key technique in solving such equations. By applying this property, we transform the equation into a simpler, more manageable form.

Step 3: Simplify the Equation

Continuing from the previous step, we have:

x+15=(35)35x + 15 = (3^5)^{\frac{3}{5}}

Using the power of a power rule, we multiply the exponents on the right side: 535=35 * \frac{3}{5} = 3. So, the equation becomes:

x+15=33x + 15 = 3^3

Now, we simplify 333^3, which is 333=273 * 3 * 3 = 27. Thus, our equation is now:

x+15=27x + 15 = 27

This simplification is significant because it transforms a complex equation with fractional exponents into a simple linear equation. The simplification process demonstrates the power of exponent rules in making equations more tractable. By applying these rules correctly, we reduce the complexity of the problem, setting the stage for the final step of isolating x. This step highlights the importance of mastering basic algebraic manipulations in solving more advanced problems.

Step 4: Isolate x

To isolate x, we subtract 15 from both sides of the equation:

x+1515=2715x + 15 - 15 = 27 - 15

This simplifies to:

x=12x = 12

So, the solution to the equation is x=12x = 12. This is the pivotal moment where we find the value of x that satisfies the original equation. Isolating x is a fundamental algebraic technique, and this step demonstrates its importance in solving equations. By performing the subtraction on both sides, we effectively separate x from the other terms, revealing its value. This direct approach underscores the elegance and efficiency of algebraic manipulations in solving mathematical problems.

Verification

It's always a good practice to verify our solution by plugging it back into the original equation. Let's substitute x=12x = 12 into (x+15)53=243(x+15)^{\frac{5}{3}} = 243:

(12+15)53=243(12 + 15)^{\frac{5}{3}} = 243

(27)53=243(27)^{\frac{5}{3}} = 243

Now, we can rewrite 27 as 333^3:

(33)53=243(3^3)^{\frac{5}{3}} = 243

Using the power of a power rule, we multiply the exponents: 353=53 * \frac{5}{3} = 5. So, we have:

35=2433^5 = 243

243=243243 = 243

Since the left side equals the right side, our solution x=12x = 12 is correct. This verification step is crucial because it confirms the accuracy of our solution and helps catch any errors made during the solving process. By substituting the solution back into the original equation, we ensure that it satisfies the equation's conditions. This practice instills confidence in our answer and reinforces the importance of careful and thorough problem-solving.

Conclusion

We have successfully solved the equation (x+15)53=243(x+15)^{\frac{5}{3}} = 243 by breaking it down into manageable steps. We first understood the meaning of fractional exponents, then isolated the term with the exponent, raised both sides to the reciprocal power, simplified the equation, and finally, isolated x. We also verified our solution to ensure its accuracy. The solution to the equation is x=12x = 12. Remember, practice makes perfect, so keep working on similar problems to strengthen your understanding of fractional exponents and equation-solving techniques. Solving equations like this one builds a strong foundation in algebra and enhances your problem-solving skills. By mastering these techniques, you'll be well-equipped to tackle more complex mathematical challenges in the future. This step-by-step approach is not just about finding the answer; it's about developing a systematic way of thinking that is applicable to a wide range of problems.

For further exploration and practice on similar mathematical concepts, you can visit Khan Academy's Algebra I section.

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