Solving The Equation (x-1)^(-4/3) = 1/81 A Step-by-Step Guide

In this article, we will walk through the step-by-step solution to the equation $(x-1)^{-\frac{4}{3}} = \frac{1}{81}$, where $x$ is a positive number. This problem involves understanding fractional exponents and algebraic manipulation. Our goal is to isolate $x$ and find its value in a simplified integer or improper fraction form. This type of equation is common in algebra and calculus, and mastering the techniques to solve it is crucial for further studies in mathematics. Understanding fractional exponents is key to tackling this problem. A fractional exponent like $-\frac{4}{3}$ indicates both a root and a power. Specifically, the denominator (3 in this case) tells us which root to take, and the numerator (4 in this case) tells us which power to raise the result to. The negative sign indicates that we need to take the reciprocal. By breaking down the exponent in this way, we can simplify the equation step by step. The algebraic manipulation skills required here include recognizing the structure of the equation, applying the same operations to both sides to maintain equality, and simplifying expressions. We'll be using properties of exponents and radicals extensively, so a good understanding of these properties is essential. Moreover, we will check our solution by substituting it back into the original equation to ensure it is correct. This step is crucial as it helps to avoid errors that can arise during the algebraic manipulation process. By the end of this article, you will not only be able to solve this specific equation but also gain a deeper understanding of how to handle similar problems involving fractional exponents and algebraic manipulations. So, let's dive in and solve this equation together, making sure to explain each step clearly and thoroughly. Remember, mathematics is like a puzzle, and each piece must fit perfectly to reveal the solution!

Step 1: Rewrite the Equation

Let's start by rewriting the given equation: $(x-1)^{-\frac{4}{3}} = \frac{1}{81}$. To deal with the negative exponent, we take the reciprocal of the base raised to the positive exponent. This gives us $\frac{1}{(x-1)^{\frac{4}{3}}} = \frac{1}{81}$. Now, our equation looks more manageable. Understanding negative exponents is crucial here. A negative exponent indicates that we need to take the reciprocal of the base. For example, $a^{-n} = \frac{1}{a^n}$. Applying this property correctly is the first step in simplifying the equation. The next important concept is understanding fractional exponents. The exponent $\frac{4}{3}$ can be interpreted as taking the cube root and then raising to the fourth power. In mathematical terms, $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$. This means that $(x-1)^{\frac{4}{3}}$ is the same as $(\sqrt[3]{x-1})^4$. By understanding this, we can rewrite the left side of the equation as $\frac{1}{(\sqrt[3]{x-1})^4}$. This transformation is key to making further progress in solving for $x$. So, our equation now looks like this: $\frac{1}{(\sqrt[3]{x-1})^4} = \frac{1}{81}$. We have successfully rewritten the equation to a more understandable form. The next step will involve eliminating the fractional exponent and simplifying further. Before moving on, it's always a good idea to pause and make sure each step is clear. Rewriting the equation is a fundamental technique in algebra, and doing it correctly sets the stage for a smooth solution process. By mastering these initial steps, we lay a solid foundation for tackling more complex problems in the future. Now, let's move on to the next step and see how we can further simplify this equation and get closer to finding the value of $x$.

Step 2: Eliminate the Fraction

To eliminate the fraction, we can take the reciprocal of both sides of the equation $\frac{1}{(\sqrt[3]{x-1})^4} = \frac{1}{81}$. Taking the reciprocal gives us $(\sqrt[3]{x-1})^4 = 81$. This step simplifies the equation by removing the fractions, making it easier to work with. The principle behind taking the reciprocal of both sides is that if two quantities are equal, their reciprocals are also equal. This is a fundamental algebraic technique that helps in clearing fractions from equations. In this case, it transforms the equation into a more manageable form where we can directly address the exponent and the radical. Now, we have an equation where the term involving $x$ is isolated on one side, which is a crucial step towards solving for $x$. Next, we need to deal with the exponent of 4. To do this, we will take the fourth root of both sides. This operation will undo the power of 4 and bring us closer to isolating the cube root term. It’s important to remember that when taking even roots, we typically consider both positive and negative solutions. However, in this particular problem, we'll focus on the positive root as the context suggests we are looking for a real solution that fits the initial equation. By applying these algebraic manipulations carefully, we are methodically simplifying the equation. Each step brings us closer to isolating $x$ and finding its value. Before we move on to the next step, let’s recap. We started with a fractional equation and, by taking reciprocals, transformed it into a simpler form. This demonstrates the power of basic algebraic techniques in problem-solving. Now, let's proceed to the next step where we will address the power of 4 and continue our journey towards finding the value of $x$.

Step 3: Take the Fourth Root

Now, we have the equation $(\sqrt[3]{x-1})^4 = 81$. To eliminate the power of 4, we take the fourth root of both sides. This gives us $\sqrt[4]{(\sqrt[3]{x-1})^4} = \sqrt[4]{81}$. Simplifying, we get $\sqrt[3]{x-1} = 3$, since $\sqrt[4]{81} = 3$. Here, understanding roots and powers is essential. Taking the fourth root is the inverse operation of raising to the power of 4. This step effectively undoes the exponent of 4, allowing us to isolate the cube root term. It's important to remember that the fourth root of 81 is indeed 3 because $3^4 = 81$. This step significantly simplifies the equation, bringing us one step closer to solving for $x$. Now, we are left with an equation involving a cube root. To eliminate the cube root, we will need to cube both sides of the equation. This is another example of using inverse operations to isolate the variable. By carefully applying these operations, we maintain the equality of the equation while making progress towards the solution. Before we move on, let's reflect on the process so far. We've rewritten the original equation, eliminated fractions, and taken a fourth root. Each step was a deliberate action to simplify the equation and bring us closer to finding the value of $x$. Now, with the cube root term isolated, we are in a good position to take the next step and solve for $x$. Let's proceed to cube both sides and continue our journey towards the solution. Remember, each step in solving an equation is like a piece of a puzzle, and we are methodically putting the pieces together.

Step 4: Cube Both Sides

To eliminate the cube root in the equation $\sqrt[3]{x-1} = 3$, we cube both sides. This means we raise both sides to the power of 3, which gives us $(\sqrt[3]{x-1})^3 = 3^3$. Simplifying, we get $x-1 = 27$. Cubing is the inverse operation of taking the cube root. By cubing both sides, we effectively undo the cube root, leaving us with a much simpler equation. In this case, $(\sqrt[3]{x-1})^3$ simplifies to $x-1$, and $3^3$ equals 27. This step is a crucial simplification that brings us very close to isolating $x$. Now, we have a straightforward linear equation: $x-1 = 27$. To solve for $x$, we simply need to add 1 to both sides of the equation. This will isolate $x$ on one side and give us its value. Before we proceed to the final step, let’s take a moment to appreciate the journey. We started with a complex equation involving fractional exponents and, through a series of algebraic manipulations, we have reduced it to a simple linear equation. This demonstrates the power of methodical problem-solving and the importance of understanding inverse operations. Now, let's move on to the final step and find the value of $x$. We are almost there, and the solution is within reach. Remember, mathematics is about breaking down complex problems into manageable steps, and we have successfully done that in this case.

Step 5: Solve for x

To solve for $x$ in the equation $x-1 = 27$, we add 1 to both sides. This gives us $x - 1 + 1 = 27 + 1$, which simplifies to $x = 28$. Therefore, the solution to the equation $(x-1)^{-\frac{4}{3}} = \frac{1}{81}$ is $x = 28$. This step is the final algebraic manipulation needed to isolate $x$ and find its value. By adding 1 to both sides, we maintain the equality of the equation while isolating $x$ on one side. The result, $x = 28$, is the solution we have been working towards throughout the entire process. Now, it's a good practice to check our solution by substituting it back into the original equation to ensure it is correct. Let's substitute $x = 28$ into the original equation: $(28-1)^{-\frac{4}{3}} = (27)^{-\frac{4}{3}}$. This can be rewritten as $\frac{1}{(27)^{\frac{4}{3}}}$. We know that $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$, so $27^{\frac{4}{3}} = (27^{\frac{1}{3}})^4 = 3^4 = 81$. Therefore, $\frac{1}{(27)^{\frac{4}{3}}} = \frac{1}{81}$, which matches the right side of the original equation. This confirms that our solution, $x = 28$, is correct. In conclusion, we have successfully solved the equation $(x-1)^{-\frac{4}{3}} = \frac{1}{81}$ by systematically applying algebraic techniques. We rewrote the equation, eliminated fractions, took the fourth root, cubed both sides, and finally, solved for $x$. We also verified our solution, ensuring its accuracy. This process demonstrates the importance of understanding fractional exponents, inverse operations, and methodical problem-solving in mathematics.

Final Answer

The solution to the equation $(x-1)^{-\frac{4}{3}} = \frac{1}{81}$ is $x = 28$.