Solving Exponential Equations 12^(x^2+5x-4) = 12^(2x+6)

Introduction

In the realm of mathematics, exponential equations play a crucial role in modeling various real-world phenomena, from population growth to radioactive decay. Exponential equations, characterized by variables in the exponents, often require specific techniques to solve. This article delves into the intricacies of solving a particular exponential equation: $12^{x^2+5x-4} = 12^{2x+6}$. We will explore the fundamental principles behind solving such equations, providing a step-by-step guide to arrive at the solution, and also discuss the importance of verifying solutions in the context of exponential functions. Understanding these concepts is not only essential for academic pursuits but also for practical applications in fields like finance, engineering, and computer science. This exploration will equip you with the tools to confidently tackle similar exponential problems and appreciate the elegance of mathematical problem-solving. By mastering the techniques discussed, you'll gain a deeper understanding of how exponential equations govern various aspects of our world. The journey through this equation will highlight the power of algebraic manipulation and the critical role of careful analysis in mathematical problem-solving.

Understanding Exponential Equations

Before we dive into the specific problem, let's establish a firm understanding of exponential equations. At its core, an exponential equation is one in which the variable appears in the exponent. These equations are distinct from polynomial equations where the variable is in the base. The fundamental principle we rely on when solving exponential equations with the same base is that if $a^m = a^n$, then $m = n$, provided that $a$ is a positive number not equal to 1. This principle stems directly from the properties of exponential functions, which are one-to-one when the base is a positive number not equal to 1. This one-to-one property ensures that each input (exponent) corresponds to a unique output (the result of the exponentiation), and vice versa. When solving exponential equations, it is crucial to manipulate the equation into a form where both sides have the same base. This often involves using properties of exponents to simplify expressions or rewriting numbers as powers of a common base. Once the bases are the same, we can equate the exponents and solve the resulting algebraic equation. It's also important to be aware of extraneous solutions, which can arise during the solving process, particularly when dealing with more complex exponential equations or equations involving radicals or logarithms. These extraneous solutions must be checked against the original equation to ensure they are valid. The elegance of solving exponential equations lies in the application of fundamental algebraic principles combined with a clear understanding of the properties of exponential functions. This combination allows us to transform seemingly complex problems into manageable algebraic equations.

Solving 12^(x2+5x-4) = 12^(2x+6) Step-by-Step

Now, let's tackle the equation $12^{x^2+5x-4} = 12^{2x+6}$ step-by-step. Our primary goal is to find the values of x that satisfy this equation.

Step 1: Equate the Exponents

Since the bases are the same (both are 12), we can equate the exponents. This is based on the fundamental property of exponential equations: if $a^m = a^n$, then $m = n$. Applying this to our equation, we get:

$x^2 + 5x - 4 = 2x + 6$

This simplifies the problem significantly, transforming it from an exponential equation into a quadratic equation, which we can solve using standard algebraic techniques. The ability to equate exponents is a powerful tool in solving exponential equations, allowing us to bypass the complexities of exponential functions and focus on the underlying algebraic relationships. However, it's crucial to remember that this step is only valid when the bases are the same and are positive numbers not equal to 1. In this case, the base is 12, which satisfies this condition. The resulting quadratic equation is a familiar form that we can manipulate and solve using methods such as factoring, completing the square, or the quadratic formula. This step highlights the interconnectedness of different areas of mathematics, as we seamlessly transition from exponential equations to quadratic equations.

Step 2: Rearrange into a Quadratic Equation

Next, we need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$. To do this, we subtract $2x$ and 6 from both sides of the equation:

$x^2 + 5x - 4 - 2x - 6 = 0$

Combining like terms, we get:

$x^2 + 3x - 10 = 0$

Now we have a standard quadratic equation ready to be solved. This rearrangement is a critical step in solving quadratic equations, as it allows us to apply well-established methods such as factoring, completing the square, or using the quadratic formula. The process of combining like terms ensures that all the terms are properly organized and grouped, making it easier to identify the coefficients a, b, and c. These coefficients are essential for applying the quadratic formula or for factoring the quadratic expression. The resulting quadratic equation, $x^2 + 3x - 10 = 0$, is a concise representation of the original exponential equation, capturing the core algebraic relationship between the variable x and the constants. This step demonstrates the importance of algebraic manipulation in simplifying equations and transforming them into solvable forms.

Step 3: Solve the Quadratic Equation

We can solve the quadratic equation $x^2 + 3x - 10 = 0$ by factoring. We are looking for two numbers that multiply to -10 and add up to 3. These numbers are 5 and -2. So, we can factor the quadratic as:

$(x + 5)(x - 2) = 0$

This gives us two possible solutions for x:

$x + 5 = 0$ or $x - 2 = 0$

Solving these linear equations, we find:

$x = -5$ or $x = 2$

Therefore, the solutions to the quadratic equation are x = -5 and x = 2. Factoring is a powerful technique for solving quadratic equations, particularly when the roots are integers or simple fractions. It involves expressing the quadratic expression as a product of two linear factors, which can then be easily solved by setting each factor equal to zero. The ability to factor quadratic expressions efficiently is a valuable skill in algebra, as it provides a direct path to finding the roots of the equation. In this case, the factors $(x + 5)$ and $(x - 2)$ clearly reveal the solutions x = -5 and x = 2. If factoring is not straightforward, alternative methods such as completing the square or using the quadratic formula can be employed. The quadratic formula, in particular, is a general method that can be used to solve any quadratic equation, regardless of the nature of its roots. This step showcases the elegance of factoring as a method for solving quadratic equations.

Step 4: Verify the Solutions

It's always a good practice to verify the solutions by substituting them back into the original equation. Let's check x = -5:

$12^{(-5)^2 + 5(-5) - 4} = 12^{2(-5) + 6}$ $12^{25 - 25 - 4} = 12^{-10 + 6}$ $12^{-4} = 12^{-4}$

The solution x = -5 is valid. Now let's check x = 2:

$12^{(2)^2 + 5(2) - 4} = 12^{2(2) + 6}$ $12^{4 + 10 - 4} = 12^{4 + 6}$ $12^{10} = 12^{10}$

The solution x = 2 is also valid. Verification is a crucial step in solving any equation, but it is particularly important when dealing with exponential equations, as extraneous solutions can sometimes arise. Extraneous solutions are solutions that are obtained through the solving process but do not satisfy the original equation. These solutions can occur due to various algebraic manipulations, such as squaring both sides of an equation or taking logarithms. By substituting the solutions back into the original equation, we can ensure that they are valid and that no errors were made during the solving process. In this case, both x = -5 and x = 2 satisfy the original equation, confirming that they are indeed the correct solutions. This step underscores the importance of rigor and attention to detail in mathematical problem-solving.

Conclusion

In conclusion, the solutions to the equation $12^{x^2+5x-4} = 12^{2x+6}$ are x = -5 and x = 2. We arrived at these solutions by applying the fundamental principle of exponential equations: if the bases are the same, then the exponents must be equal. This allowed us to transform the exponential equation into a quadratic equation, which we then solved by factoring. Finally, we verified our solutions by substituting them back into the original equation. This comprehensive approach highlights the key steps in solving exponential equations and the importance of verifying solutions. The process of solving this equation demonstrates the interconnectedness of different mathematical concepts, from exponential functions to quadratic equations. It also reinforces the importance of algebraic manipulation, factoring, and verification in mathematical problem-solving. By mastering these techniques, you can confidently tackle a wide range of exponential equations and appreciate the elegance and power of mathematical methods. This journey through solving a single equation provides valuable insights into the broader field of exponential functions and their applications in various disciplines. The ability to solve exponential equations is a fundamental skill in mathematics and is essential for understanding and modeling various real-world phenomena.