Solving Exponential Equations Using The One-to-One Property

In the realm of mathematics, exponential equations hold a significant place, particularly in modeling various real-world phenomena such as population growth, radioactive decay, and compound interest. These equations involve variables in the exponents, making them distinct from polynomial equations. Among the techniques for solving exponential equations, the one-to-one property stands out as a powerful tool when dealing with equations where the bases are the same. This property simplifies the process of solving for the unknown variable by allowing us to equate the exponents directly. This article delves into the application of the one-to-one property in solving exponential equations, using a specific example to illustrate the method. We will explore the underlying principle of the one-to-one property, demonstrate its application through a step-by-step solution, and discuss the broader implications and advantages of using this technique. Mastering the one-to-one property not only enhances your problem-solving skills in algebra but also provides a solid foundation for more advanced mathematical concepts.

Before we dive into solving the equation, it's crucial to understand the one-to-one property of exponential functions. This property states that if $a^m = a^n$, where $a$ is a positive number not equal to 1, then $m = n$. In simpler terms, if two exponential expressions with the same base are equal, then their exponents must also be equal. This property stems from the fact that exponential functions are one-to-one, meaning that each input corresponds to a unique output, and vice versa. This uniqueness allows us to equate the exponents and solve for the variable. The condition that $a$ must be a positive number not equal to 1 is essential because if $a$ were 1, then $1^m = 1^n$ would be true for any values of $m$ and $n$, making the property useless. Similarly, if $a$ were 0 or negative, the exponential function would not be well-defined for all exponents, and the one-to-one property would not hold. The one-to-one property significantly simplifies solving exponential equations because it transforms a potentially complex equation into a simpler algebraic equation involving only the exponents. This simplification is particularly useful when dealing with equations where the exponents are expressions involving the variable we want to solve for. By understanding and applying this property, we can efficiently find the solutions to a wide range of exponential equations.

Now, let's apply the one-to-one property to solve the given equation: $e^{x^2 + 20} = e^{9x}$. This equation presents a classic scenario where the one-to-one property can be effectively used. We have two exponential expressions with the same base, $e$, which is the base of the natural exponential function (approximately 2.71828). The exponents are $x^2 + 20$ and $9x$, respectively. Our goal is to find the values of $x$ that satisfy this equation. The equation's structure immediately suggests the application of the one-to-one property. Since the bases are the same, we can equate the exponents and transform the exponential equation into a quadratic equation, which is a familiar type of equation to solve. This step is crucial because it allows us to move from the realm of exponential functions to the more manageable world of polynomial equations. By solving the resulting quadratic equation, we will find the values of $x$ that make the original exponential equation true. The problem statement clearly sets the stage for using the one-to-one property, and the subsequent steps will demonstrate how this property is applied to arrive at the solution. The transition from an exponential equation to a quadratic equation highlights the power and utility of the one-to-one property in simplifying complex mathematical problems.

Step-by-Step Solution

To solve the equation $e^{x^2 + 20} = e^{9x}$, we will follow these steps:

1. Apply the One-to-One Property: Since the bases are the same (both are $e$), we can equate the exponents:

   $x^2 + 20 = 9x$

   This step is the direct application of the one-to-one property, which is the cornerstone of our solution. By equating the exponents, we transform the exponential equation into a more manageable algebraic equation. This transformation is crucial because it allows us to use standard algebraic techniques to solve for the unknown variable. The one-to-one property provides a bridge between exponential equations and polynomial equations, making the problem more accessible.

2. Rearrange the Equation: To solve for $x$, we need to rearrange the equation into a standard quadratic form, which is $ax^2 + bx + c = 0$. Subtract $9x$ from both sides:

   $x^2 - 9x + 20 = 0$

   This step involves simple algebraic manipulation to bring all terms to one side of the equation, leaving zero on the other side. This rearrangement is necessary to apply standard methods for solving quadratic equations, such as factoring or using the quadratic formula. The standard quadratic form is essential for identifying the coefficients $a$, $b$, and $c$, which are needed for these solution methods. Rearranging the equation into the standard form is a crucial step in the process of solving quadratic equations and is a common practice in algebra.

3. Solve the Quadratic Equation: The quadratic equation $x^2 - 9x + 20 = 0$ can be solved by factoring. We look for two numbers that multiply to 20 and add up to -9. These numbers are -4 and -5. So, we can factor the equation as:

   $(x - 4)(x - 5) = 0$

   Factoring is a powerful technique for solving quadratic equations, especially when the roots are integers or simple fractions. It involves expressing the quadratic expression as a product of two linear factors. In this case, we found that the quadratic expression $x^2 - 9x + 20$ can be factored into $(x - 4)(x - 5)$. This factorization directly leads to the solutions for $x$ by setting each factor equal to zero. The ability to factor a quadratic equation significantly simplifies the process of finding its roots. The numbers -4 and -5 were identified by considering the factors of the constant term (20) and their sums, which must equal the coefficient of the linear term (-9). This systematic approach to factoring ensures that we find the correct factors.

4. Find the Solutions: Set each factor equal to zero and solve for $x$:

   $x - 4 = 0$ or $x - 5 = 0$

   $x = 4$ or $x = 5$

   Setting each factor equal to zero is a direct consequence of the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This property is fundamental to solving equations by factoring. By setting each factor $(x - 4)$ and $(x - 5)$ equal to zero, we create two simple linear equations that can be easily solved for $x$. The solutions $x = 4$ and $x = 5$ are the values of $x$ that make the original quadratic equation true, and therefore, they are also the solutions to the original exponential equation. This step highlights the final stage in the process of solving an equation by factoring, where the factors are used to determine the roots of the equation.

Therefore, the solutions for $x$ are 4 and 5. We can write the answer as a comma-separated list:

$x = 4, 5$

These values of $x$ satisfy the original equation $e^{x^2 + 20} = e^{9x}$. To verify this, we can substitute each value back into the original equation:

For $x = 4$:

$e^{(4^2 + 20)} = e^{(16 + 20)} = e^{36}$

$e^{(9 imes 4)} = e^{36}$

For $x = 5$:

$e^{(5^2 + 20)} = e^{(25 + 20)} = e^{45}$

$e^{(9 imes 5)} = e^{45}$

Both values satisfy the equation, confirming that our solutions are correct. This verification step is crucial in ensuring the accuracy of the solutions and demonstrating a thorough understanding of the problem-solving process. By substituting the solutions back into the original equation, we can confirm that the left-hand side equals the right-hand side for each solution. This step provides confidence in the correctness of the answers and reinforces the understanding of how the solutions relate to the original equation. The final answer, presented clearly and concisely, is the culmination of the problem-solving process, showcasing the application of the one-to-one property and the techniques for solving quadratic equations.

In conclusion, the one-to-one property is a valuable tool for solving exponential equations where the bases are the same. By equating the exponents, we can transform a complex exponential equation into a simpler algebraic equation, such as a quadratic equation in this case. The step-by-step solution presented demonstrates how to apply this property effectively, leading to the solutions $x = 4$ and $x = 5$. The one-to-one property simplifies the process of solving exponential equations and is a fundamental concept in algebra. This technique is not only applicable to simple equations but also extends to more complex scenarios where the exponents involve more intricate expressions. Mastering this property enhances one's ability to solve a wide range of exponential equations, making it an essential tool in mathematical problem-solving. The ability to transform exponential equations into algebraic equations through the one-to-one property highlights the interconnectedness of different mathematical concepts and techniques. This approach underscores the importance of understanding fundamental principles and their applications in various mathematical contexts. The example provided serves as a clear illustration of the one-to-one property's effectiveness and its role in simplifying and solving exponential equations. The process of verifying the solutions further reinforces the importance of accuracy and thoroughness in mathematical problem-solving.