Solving Exponential Equations Find X In X^(-6/7) = 64

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In this article, we will delve into solving the equation xβˆ’67=64x^{-\frac{6}{7}} = 64 for xx, assuming xx is positive. This problem falls under the category of exponential equations, where the variable appears in the exponent. Solving such equations often involves using the properties of exponents and logarithms, but in this particular case, we can solve it by manipulating the equation algebraically. We will explore each step in detail to ensure a clear understanding of the solution process. Our goal is to express xx as a simplified integer or an improper fraction, if necessary. Let's embark on this mathematical journey and unravel the value of xx in this intriguing equation.

Before we dive into the solution, it's essential to grasp the fundamentals of exponential equations. An exponential equation is an equation in which a variable occurs in the exponent. These equations often require us to use properties of exponents and logarithms to isolate the variable. In our case, we have xβˆ’67=64x^{-\frac{6}{7}} = 64. The negative exponent and the fractional exponent both provide clues on how to approach the problem. A negative exponent indicates a reciprocal, and a fractional exponent represents a root and a power. Understanding these properties is crucial for effectively solving this equation.

Key Concepts in Exponential Equations

  1. Negative Exponents: A term raised to a negative power is equivalent to the reciprocal of the term raised to the positive power. For example, aβˆ’n=1ana^{-n} = \frac{1}{a^n}.
  2. Fractional Exponents: A fractional exponent can be interpreted as a root and a power. For instance, amna^{\frac{m}{n}} can be written as amn\sqrt[n]{a^m} or (an)m(\sqrt[n]{a})^m.
  3. Inverse Operations: To solve for a variable in an exponent, we often use inverse operations. The inverse operation of exponentiation is taking a root or logarithm.

In the context of our equation, xβˆ’67=64x^{-\frac{6}{7}} = 64, we will use these concepts to isolate xx and find its value. The negative exponent suggests we will be dealing with reciprocals, and the fractional exponent indicates we need to consider both roots and powers. By applying these principles systematically, we can solve for xx and express it in its simplest form.

Now, let's proceed with the step-by-step solution of the equation xβˆ’67=64x^{-\frac{6}{7}} = 64. We will break down each step to ensure clarity and understanding.

Step 1: Eliminate the Negative Exponent

Our first step is to eliminate the negative exponent. Recall that a negative exponent means we take the reciprocal of the base raised to the positive exponent. Thus, we rewrite the equation as:

1x67=64\frac{1}{x^{\frac{6}{7}}} = 64

This transformation makes the exponent positive, which is easier to work with. By taking the reciprocal, we've set the stage for further simplification.

Step 2: Isolate the Term with the Fractional Exponent

Next, we want to isolate the term with the fractional exponent. To do this, we can take the reciprocal of both sides of the equation. This gives us:

x67=164x^{\frac{6}{7}} = \frac{1}{64}

Now, we have xx raised to a positive fractional exponent isolated on one side of the equation. This is a significant step forward, as we can now address the fractional exponent directly.

Step 3: Eliminate the Fractional Exponent

To eliminate the fractional exponent, we raise both sides of the equation to the reciprocal of the exponent. In this case, the exponent is 67\frac{6}{7}, so we raise both sides to the power of 76\frac{7}{6}:

(x67)76=(164)76(x^{\frac{6}{7}})^{\frac{7}{6}} = (\frac{1}{64})^{\frac{7}{6}}

Using the power of a power rule, which states that (am)n=amn(a^m)^n = a^{mn}, we simplify the left side:

x(67β‹…76)=x1=xx^{(\frac{6}{7} \cdot \frac{7}{6})} = x^1 = x

So, we have:

x=(164)76x = (\frac{1}{64})^{\frac{7}{6}}

This step is crucial because it isolates xx, allowing us to find its value by simplifying the right side of the equation.

Step 4: Simplify the Right Side

Now, we need to simplify (164)76(\frac{1}{64})^{\frac{7}{6}}. Recall that a fractional exponent represents both a root and a power. The denominator of the fraction (6) indicates the root, and the numerator (7) indicates the power. So, we can rewrite the expression as:

x=(1646)7x = (\sqrt[6]{\frac{1}{64}})^7

First, let's find the sixth root of 164\frac{1}{64}. We know that 64=2664 = 2^6, so 164=126\frac{1}{64} = \frac{1}{2^6}. Therefore:

1646=1266=12\sqrt[6]{\frac{1}{64}} = \sqrt[6]{\frac{1}{2^6}} = \frac{1}{2}

Now, we raise this result to the power of 7:

x=(12)7=127=1128x = (\frac{1}{2})^7 = \frac{1}{2^7} = \frac{1}{128}

Thus, we have found the value of xx.

The solution to the equation xβˆ’67=64x^{-\frac{6}{7}} = 64 is:

x=1128x = \frac{1}{128}

This is a simplified improper fraction. We have successfully solved for xx by systematically applying the properties of exponents and roots. Each step was carefully executed to ensure accuracy and clarity. The final result, 1128\frac{1}{128}, represents the value of xx that satisfies the original equation.

Besides the step-by-step solution outlined above, there's an alternative method to solve the equation xβˆ’67=64x^{-\frac{6}{7}} = 64. This method involves converting 64 into its exponential form with a base related to the exponent in the equation. Let’s explore this alternative approach.

Step 1: Express 64 as a Power of 2

First, we express 64 as a power of 2. We know that 64=2664 = 2^6. Thus, we can rewrite the original equation as:

xβˆ’67=26x^{-\frac{6}{7}} = 2^6

This step is crucial because it sets the stage for equating the exponents after adjusting the base.

Step 2: Raise Both Sides to the Power of βˆ’76-\frac{7}{6}

To isolate xx, we raise both sides of the equation to the power of βˆ’76-\frac{7}{6}, which is the reciprocal of βˆ’67-\frac{6}{7}:

(xβˆ’67)βˆ’76=(26)βˆ’76(x^{-\frac{6}{7}})^{-\frac{7}{6}} = (2^6)^{-\frac{7}{6}}

Using the power of a power rule, (am)n=amn(a^m)^n = a^{mn}, we simplify both sides:

x^{(-\frac{6}{7} \cdot -\frac{7}{6})} = 2^{(6 \cdot - rac{7}{6})}

Simplifying the exponents, we get:

x1=2βˆ’7x^1 = 2^{-7}

Thus,

x=2βˆ’7x = 2^{-7}

This step is a direct application of the power rule, making it easier to isolate xx.

Step 3: Simplify the Result

Finally, we simplify 2βˆ’72^{-7}. Recall that a negative exponent indicates a reciprocal, so:

x=2βˆ’7=127x = 2^{-7} = \frac{1}{2^7}

We know that 27=1282^7 = 128, so:

x=1128x = \frac{1}{128}

This is the same result we obtained using the step-by-step method. This alternative approach provides a more direct route to the solution by leveraging the exponential form of 64.

In this article, we successfully solved the equation xβˆ’67=64x^{-\frac{6}{7}} = 64 using two different methods. Both approaches led us to the same solution: x=1128x = \frac{1}{128}. The first method involved systematically eliminating the negative and fractional exponents, while the second method utilized the exponential form of 64 to simplify the equation. Understanding these different approaches enhances our problem-solving skills and provides us with flexibility when tackling similar exponential equations. The solution, 1128\frac{1}{128}, is a simplified improper fraction, fulfilling the requirements of the problem statement. Through this detailed exploration, we have not only solved the equation but also reinforced key concepts in algebra and exponential functions. Solving equations like this requires a solid understanding of exponential and root properties. Whether you prefer the step-by-step method or the alternative approach, the key is to apply these properties correctly and consistently. The ability to manipulate exponents and roots is a valuable skill in mathematics, and this problem serves as an excellent exercise in honing that skill. By practicing these techniques, you'll be well-equipped to handle more complex equations in the future. The world of mathematics is filled with intricate problems, but with a clear understanding of the underlying principles, we can confidently navigate and solve them. This exploration of exponential equations is just one step in that journey, and the more we practice, the more proficient we become.