Solve The Exponential Equation 9^x - 81 = 3^x + 1/27.

Jun 19, 2025 by ADMIN 54 views

Solving the Exponential Equation 9^x - 81 = 3^x + 1/27: A Comprehensive Guide

Introduction

In this article, we will delve into the process of solving the exponential equation 9^x - 81 = 3^x + 1/27. Exponential equations, which involve variables in the exponents, often require specific techniques to find their solutions. This particular equation combines exponential terms with different bases, making it an interesting challenge to tackle. Our journey will involve transforming the equation into a more manageable form, applying algebraic manipulations, and ultimately determining the value(s) of x that satisfy the equation. Understanding how to solve such equations is crucial in various fields, including mathematics, physics, engineering, and computer science, where exponential models are frequently used to describe real-world phenomena. By the end of this guide, you will have a clear understanding of the steps involved and the reasoning behind them, equipping you with the skills to solve similar exponential equations.

Understanding Exponential Equations

Before we dive into the specifics of our equation, let's briefly discuss exponential equations in general. An exponential equation is one in which the variable appears in the exponent. The key to solving these equations often lies in manipulating them so that we can compare exponents or use logarithms. Exponential functions, represented as a^x, where a is the base and x is the exponent, are fundamental in mathematics and appear in numerous applications. The behavior of these functions depends heavily on the base a. If a is greater than 1, the function represents exponential growth, while if a is between 0 and 1, it represents exponential decay. When solving equations involving exponentials, it’s essential to look for ways to simplify the expressions and find common bases. This often involves using properties of exponents, such as a^(m+n) = a^m * a^n and (a^m)n = a^(mn)*. Recognizing these properties and applying them strategically can significantly simplify the solution process. Moreover, understanding the relationship between exponential and logarithmic functions is crucial, as logarithms are the inverse of exponentials and can be used to solve equations where isolating the variable in the exponent is necessary. In the following sections, we’ll apply these concepts to solve the given equation, demonstrating how a methodical approach can lead to a clear solution.

Transforming the Equation

The first step in solving the exponential equation 9^x - 81 = 3^x + 1/27 is to transform it into a more manageable form. We notice that 9 and 81 are powers of 3. Specifically, 9 = 3^2 and 81 = 3^4. Also, 1/27 can be expressed as 3^(-3). Rewriting the equation using these powers of 3 will help us to consolidate the exponential terms and simplify the overall structure. By expressing all terms with a common base, we can more easily compare the exponents and potentially apply algebraic techniques to solve for x. This initial transformation is a critical step because it sets the stage for subsequent simplifications. Without this step, the equation would remain in a complex form that is difficult to work with. The ability to recognize these relationships between numbers and express them in terms of a common base is a fundamental skill in solving exponential equations. It allows us to convert complex expressions into simpler, more manageable forms, paving the way for the application of other algebraic techniques. This process of rewriting the equation also highlights the importance of understanding the properties of exponents and how they can be used to manipulate expressions. Let's proceed with this transformation to make the equation more approachable.

Rewriting with Base 3

Now, let's rewrite the equation 9^x - 81 = 3^x + 1/27 using base 3. We know that 9 = 3^2, 81 = 3^4, and 1/27 = 3^(-3). Substituting these into the original equation, we get (3²⁾x - 3^4 = 3^x + 3^(-3). Applying the power of a power rule, which states that (a^m)n = a^(mn)*, we can simplify (3²⁾x to 3^(2x). Our equation now looks like this: 3^(2x) - 3^4 = 3^x + 3^(-3). This transformation is crucial because it brings all the terms to a common base, allowing us to treat them in a more uniform manner. By having a common base, we can start to think about the relationships between the exponents and how we might isolate the variable x. The equation is now in a form where we can see the exponential terms more clearly and consider algebraic manipulations that might help us solve for x. This step is a testament to the power of recognizing patterns and using fundamental properties of exponents to simplify complex equations. The ability to convert expressions to a common base is a cornerstone technique in solving exponential equations, and this example demonstrates its effectiveness in transforming a seemingly complex equation into a more manageable one. Let's move on to further simplifications now that we have a common base.

Rearranging the Equation

With the equation now in the form 3^(2x) - 3^4 = 3^x + 3^(-3), the next step is to rearrange the equation to group similar terms together. This often involves moving all terms to one side of the equation, resulting in an expression equal to zero. Doing so can reveal underlying structures or patterns that are not immediately apparent in the original form. By bringing all terms to one side, we set the stage for potential factorization or other algebraic manipulations that can help us isolate the variable x. This rearrangement is not just a matter of aesthetics; it’s a strategic move to put the equation in a format that is conducive to solving. In this particular case, we'll subtract 3^x and 3^(-3) from both sides to get all terms on the left-hand side. This process is a common technique in algebra and is essential for solving various types of equations, not just exponential ones. The goal is to create an expression that is easier to analyze and manipulate, ultimately leading to a solution. This step demonstrates the importance of algebraic manipulation in simplifying equations and making them more solvable. Let’s proceed with this rearrangement and see what form our equation takes.

Setting Up a Quadratic Form

After rearranging the equation, we have 3^(2x) - 3^x - 3^4 - 3^(-3) = 0. Now, we can recognize that this equation can be viewed as a quadratic form. To see this more clearly, let's make a substitution. Let y = 3^x. Then, y^2 = (3^x)2 = 3^(2x). Substituting y into the equation, we get y^2 - y - 3^4 - 3^(-3) = 0. This equation is now in the form of a quadratic equation, specifically y^2 - y - (81 + 1/27) = 0. Recognizing this quadratic form is a crucial step because it allows us to apply techniques that are well-established for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. This transformation is a powerful example of how recognizing underlying structures in equations can lead to simpler solutions. By making the substitution y = 3^x, we have effectively converted a complex exponential equation into a familiar quadratic equation. This ability to transform equations into more recognizable forms is a key skill in mathematical problem-solving. It often involves looking beyond the surface appearance of an equation and identifying patterns that can be exploited. Now that we have a quadratic equation, we can proceed to solve for y using standard methods.

Solving the Quadratic Equation

Now that we have the quadratic equation y^2 - y - (81 + 1/27) = 0, let's solve for y. We can use the quadratic formula, which states that for an equation of the form ay^2 + by + c = 0, the solutions for y are given by:

y = [-b ± sqrt(b^2 - 4ac)] / (2a)

In our case, a = 1, b = -1, and c = -(81 + 1/27). Plugging these values into the quadratic formula, we get:

y = [1 ± sqrt((-1)^2 - 4 * 1 * (-81 - 1/27))] / (2 * 1)

Simplifying the expression under the square root:

(-1)^2 - 4 * 1 * (-81 - 1/27) = 1 + 4 * (81 + 1/27) = 1 + 324 + 4/27 = 325 + 4/27 = (325 * 27 + 4) / 27 = (8775 + 4) / 27 = 8779 / 27

So, we have:

y = [1 ± sqrt(8779 / 27)] / 2

This gives us two possible values for y:

y_1 = [1 + sqrt(8779 / 27)] / 2

y_2 = [1 - sqrt(8779 / 27)] / 2

Solving the quadratic equation is a crucial step in finding the solutions to our original exponential equation. The quadratic formula is a powerful tool that allows us to find the roots of any quadratic equation, regardless of whether it can be factored easily. In this case, the quadratic formula provides us with two potential values for y, which we will then use to find the corresponding values for x. This process highlights the interconnectedness of different areas of mathematics, as we are using techniques from quadratic equations to solve an exponential equation. Now that we have the values for y, we need to go back to our original substitution and find the values for x that correspond to these values of y. This will involve using the relationship y = 3^x and solving for x, which we will do in the next section.

Back-Substitution to Find x

Now that we have the values for y, which are y_1 = [1 + sqrt(8779 / 27)] / 2 and y_2 = [1 - sqrt(8779 / 27)] / 2, we need to back-substitute to find the corresponding values for x. Recall that we made the substitution y = 3^x. Therefore, we need to solve the equations 3^x = y_1 and 3^x = y_2 for x. To solve for x in these exponential equations, we can use logarithms. Taking the logarithm base 3 of both sides, we get:

x = log_3(y_1) and x = log_3(y_2)

Let's first consider y_1 = [1 + sqrt(8779 / 27)] / 2. Since y_1 is a positive value, we can find a real solution for x. Using the change of base formula for logarithms, we can express log_3(y_1) in terms of natural logarithms (ln) or common logarithms (log base 10):

x_1 = log_3(y_1) = ln(y_1) / ln(3)

Now, let's consider y_2 = [1 - sqrt(8779 / 27)] / 2. Since sqrt(8779 / 27) is significantly greater than 1, y_2 will be a negative value. However, 3^x is always positive for any real number x. Therefore, there is no real solution for x corresponding to y_2. This means that we only have one real solution for our original equation, which comes from y_1. Back-substitution is a critical step in solving equations that involve substitutions. It allows us to relate the solutions of the substituted equation back to the original variables. In this case, it enabled us to find the value of x that satisfies the original exponential equation. This process also highlights the importance of checking the validity of solutions. We found that one of the values of y led to no real solution for x, which is a common occurrence in solving exponential and logarithmic equations. We've now found the real solution for x, which is x_1 = log_3(y_1) = ln(y_1) / ln(3), where y_1 = [1 + sqrt(8779 / 27)] / 2. This value represents the solution to the exponential equation 9^x - 81 = 3^x + 1/27.

Final Solution

After performing all the steps, we have arrived at the final solution for the exponential equation 9^x - 81 = 3^x + 1/27. We found that there is only one real solution, which is given by:

x = log_3([1 + sqrt(8779 / 27)] / 2)

This can also be expressed using natural logarithms as:

x = ln([1 + sqrt(8779 / 27)] / 2) / ln(3)

This solution represents the value of x that satisfies the original equation. To summarize the process, we started by transforming the equation to a common base, then rearranged it into a quadratic form using a substitution. We solved the quadratic equation using the quadratic formula, which gave us two potential values for the substituted variable. We then back-substituted to find the corresponding values for x. We found that one of the values led to no real solution, while the other provided us with the real solution for x. The process of solving this exponential equation demonstrates the importance of several key mathematical techniques, including recognizing patterns, transforming equations, using substitutions, and applying the quadratic formula. It also highlights the need to check the validity of solutions, as not all solutions obtained through algebraic manipulation may be valid in the original equation. This comprehensive approach ensures that we arrive at the correct solution and understand the underlying principles involved. The final solution, x = log_3([1 + sqrt(8779 / 27)] / 2), is the answer to our problem and showcases the power of combining various algebraic and logarithmic techniques to solve complex equations.

Conclusion

In conclusion, solving the exponential equation 9^x - 81 = 3^x + 1/27 involved a multi-step process that combined algebraic manipulation, substitution, and the application of logarithmic properties. We began by transforming the equation to a common base, which allowed us to simplify the terms and recognize a quadratic form. The substitution technique was crucial in converting the exponential equation into a more manageable quadratic equation. We then used the quadratic formula to find the solutions for the substituted variable, and back-substitution helped us determine the corresponding values for x. Along the way, we emphasized the importance of checking the validity of solutions to ensure they satisfy the original equation. The final solution, x = log_3([1 + sqrt(8779 / 27)] / 2), represents the real value of x that makes the equation true. This exercise underscores the significance of a systematic approach to problem-solving in mathematics. By breaking down a complex equation into smaller, more manageable steps, we were able to apply various techniques and arrive at the solution. The skills and concepts demonstrated in this solution are valuable in a wide range of mathematical and scientific applications. Understanding how to solve exponential equations is not only important for academic purposes but also for practical applications in fields such as physics, engineering, and finance, where exponential models are frequently used to describe real-world phenomena. This comprehensive guide provides a clear and detailed explanation of the solution process, equipping readers with the knowledge and skills to tackle similar exponential equations with confidence. By mastering these techniques, you can enhance your problem-solving abilities and deepen your understanding of mathematical principles.