Solving (x-3)^(-1/3) = 1/5 A Step-by-Step Guide

Introduction

In this article, we will delve into the step-by-step solution of the equation $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$. This equation involves a fractional exponent, which can initially seem daunting. However, by applying the fundamental principles of algebra and exponent manipulation, we can methodically isolate x and arrive at the simplified integer or improper fraction solution. Understanding these steps is vital for anyone studying algebra or preparing for standardized mathematics tests. This comprehensive guide will not only provide the answer but also explain the underlying logic, ensuring a thorough grasp of the methods involved. We'll explore how negative exponents indicate reciprocals and how fractional exponents relate to roots, providing a solid foundation for solving similar problems in the future. By the end of this discussion, you'll be equipped with the knowledge to tackle equations of this nature with confidence.

Understanding the Equation

Before diving into the solution, let’s break down the equation $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$. The left-hand side features a term raised to a negative fractional exponent. Recall that a negative exponent signifies the reciprocal of the base raised to the positive exponent. In other words, $a^{-n} = \frac{1}{a^n}$. Applying this principle, we can rewrite $(x-3)^{-\frac{1}{3}}$ as $\frac{1}{(x-3)^{\frac{1}{3}}}$. The fractional exponent $\frac{1}{3}$ indicates the cube root. Therefore, the term $(x-3)^{\frac{1}{3}}$ is equivalent to $\sqrt[3]{x-3}$. Substituting these interpretations, our equation transforms to $\frac{1}{\sqrt[3]{x-3}}=\frac{1}{5}$. This transformation clarifies the structure of the equation, making it easier to visualize and manipulate. We are essentially looking for a value of x such that the cube root of (x-3) is equal to 5. This understanding is crucial as it guides our subsequent steps in solving for x. The emphasis on understanding each component of the equation is critical for developing problem-solving skills in algebra.

Step-by-Step Solution

Now, let's systematically solve the equation $\frac{1}{\sqrt[3]{x-3}}=\frac{1}{5}$.

1. Take the reciprocal of both sides: To eliminate the fractions, we take the reciprocal of both sides of the equation. The reciprocal of $\frac{1}{\sqrt[3]{x-3}}$ is $\sqrt[3]{x-3}$, and the reciprocal of $\frac{1}{5}$ is 5. Thus, the equation becomes $\sqrt[3]{x-3} = 5$. This step simplifies the equation by removing the fractions, making it easier to work with.

2. Cube both sides: To eliminate the cube root, we raise both sides of the equation to the power of 3. This operation cancels out the cube root on the left side, giving us $(\sqrt[3]{x-3})^3 = 5^3$. Simplifying this, we get $x-3 = 125$. Cubing both sides is a crucial step in isolating x and removing the radical.

3. Isolate x: To isolate x, we add 3 to both sides of the equation. This gives us $x - 3 + 3 = 125 + 3$, which simplifies to $x = 128$. Adding 3 to both sides effectively moves the constant term to the right side of the equation, leaving x by itself on the left.

4. Check the solution: It's always good practice to check our solution by substituting x = 128 back into the original equation: $(128-3)^{-\frac{1}{3}} = (125)^{-\frac{1}{3}} = \frac{1}{(125)^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{125}} = \frac{1}{5}$. Since this matches the right-hand side of the original equation, our solution is correct. This verification step ensures that no errors were made during the solving process.

Therefore, the solution to the equation $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$ is $x = 128$.

Alternative Approaches

While the step-by-step method described above is straightforward, there are alternative approaches to solving the equation $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$. Understanding these different methods can enhance your problem-solving skills and provide flexibility when tackling similar problems.

Direct Manipulation of Exponents

Instead of taking reciprocals and cubing separately, we can directly manipulate the exponents. Starting with $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$, we can rewrite $\frac{1}{5}$ as $5^{-1}$. This gives us $(x-3)^{-\frac{1}{3}} = 5^{-1}$. To eliminate the negative fractional exponent on the left side, we can raise both sides to the power of -3. This gives us $((x-3)^{-\frac{1}{3}})^{-3} = (5^{-1})^{-3}$. Using the property of exponents that $(a^m)^n = a^{mn}$, we simplify the left side to $x-3$ and the right side to $5^3$, resulting in $x-3 = 125$. Adding 3 to both sides then gives us $x = 128$, as before. This method directly addresses the exponents, streamlining the process into fewer steps.

Using Logarithms

Another approach involves the use of logarithms, although this method is less common for this type of problem but illustrates a versatile mathematical tool. Starting with $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$, we can take the logarithm of both sides. Let's use the natural logarithm (ln). We have $\ln((x-3)^{-\frac{1}{3}}) = \ln(\frac{1}{5})$. Using the logarithmic property $\ln(a^b) = b\ln(a)$, we rewrite the left side as $-\frac{1}{3}\ln(x-3) = \ln(\frac{1}{5})$. Multiplying both sides by -3 gives $\ln(x-3) = -3\ln(\frac{1}{5})$. Since $\ln(\frac{1}{5}) = \ln(5^{-1}) = -\ln(5)$, we have $\ln(x-3) = 3\ln(5) = \ln(5^3) = \ln(125)$. Taking the exponential of both sides to remove the natural logarithm gives $x-3 = 125$, which again leads to $x = 128$. This approach demonstrates the flexibility afforded by logarithms in solving exponential equations.

Common Mistakes to Avoid

When solving equations like $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$, several common mistakes can lead to incorrect answers. Awareness of these pitfalls is crucial for accurate problem-solving. One frequent error is mishandling negative exponents. Remember, a negative exponent indicates the reciprocal of the base raised to the positive exponent, i.e., $a^{-n} = \frac{1}{a^n}$. Forgetting this rule can lead to incorrect simplification. Another common mistake is improperly applying the fractional exponent. The exponent $\frac{1}{3}$ signifies the cube root, not the division by 3. Confusing these operations can lead to significant errors in the solution. Furthermore, errors can occur during algebraic manipulations, such as failing to apply operations to both sides of the equation equally. For instance, when cubing both sides to eliminate the cube root, ensure that both the left and right sides are cubed correctly. Another potential mistake is neglecting to check the solution. Always substitute the solution back into the original equation to verify its correctness. This step helps catch any arithmetic errors or inconsistencies. Finally, be careful with the order of operations. Ensure that you follow the correct sequence of operations (PEMDAS/BODMAS) to avoid errors in simplification. By being mindful of these common pitfalls, you can significantly improve your accuracy in solving algebraic equations.

Conclusion

In summary, we have successfully solved the equation $(x-3)^{-\frac{1}{3}}=\frac{1}{5}$ for $x$. By understanding the principles of negative and fractional exponents, applying algebraic manipulations step-by-step, and verifying the solution, we arrived at the answer $x = 128$. This problem highlights the importance of a solid grasp of exponent rules, radical operations, and algebraic techniques. We also explored alternative methods, including direct exponent manipulation and the use of logarithms, demonstrating that there can be multiple paths to the solution. Recognizing and avoiding common mistakes further ensures accuracy in problem-solving. The ability to solve equations like this is a fundamental skill in algebra and is crucial for more advanced mathematical concepts. Practice and familiarity with these techniques will build confidence and proficiency in solving a wide range of equations. Through this detailed explanation, we hope you've gained a deeper understanding of how to approach and solve equations involving exponents and roots.