Solving Exponential Equations Find The Values Of X For 25^x = 5^(x^2-3)

At the heart of this mathematical exploration lies the equation 25^x = 5^(x2-3). Our mission is to dissect this equation, employing the principles of exponents and algebraic manipulation, to unearth the precise values of x that satisfy this intriguing relationship. This equation, at first glance, might appear daunting, but with a systematic approach, we can navigate its complexities and arrive at the solutions. Before we dive into the intricacies of solving for x, it's crucial to understand the fundamental concepts that underpin this equation. We're dealing with exponential functions, where the variable x appears in the exponent. This means that the value of the expression changes exponentially as x varies. The key to solving this equation lies in recognizing the common base between the two sides. Both 25 and 5 can be expressed as powers of 5, which will allow us to equate the exponents and form a more manageable algebraic equation. The journey to find the values of x that make this equation true is a fascinating one, involving the interplay of exponents, algebraic manipulation, and a keen eye for detail. Let's embark on this mathematical quest together and unravel the mystery behind this equation.

Laying the Foundation: Expressing Both Sides with a Common Base

To effectively tackle the equation 25^x = 5^(x2-3), our initial step involves expressing both sides with a common base. The keen observer will immediately recognize that 25 is simply 5 squared, or 5^2. This crucial insight allows us to rewrite the left-hand side of the equation, 25^x, as (5²⁾x. Now, invoking the power of a power rule, which states that (a^m)n = a^(m*n), we can further simplify (5²⁾x to 5^(2x). This transformation is a cornerstone of our solution strategy, as it brings us closer to a unified expression where we can directly compare exponents. With this strategic maneuver, our equation now takes on a more streamlined appearance: 5^(2x) = 5^(x2-3). The significance of this transformation cannot be overstated. By expressing both sides of the equation with the same base (5 in this case), we've effectively leveled the playing field. We've created a scenario where the exponential functions are directly comparable, and the solutions for x will emerge from the relationship between the exponents themselves. This step is not just a matter of simplification; it's a fundamental shift in perspective that unlocks the pathway to solving the equation. The power of mathematical manipulation lies in its ability to transform complex expressions into simpler, more manageable forms. In this case, expressing both sides with a common base is the key that unlocks the door to solving for x. Now that we have a common base, we can proceed to the next stage: equating the exponents.

Equating the Exponents: A Path to a Quadratic Equation

With both sides of the equation 25^x = 5^(x2-3) now expressed with a common base, we stand at a pivotal juncture. We've transformed the equation into 5^(2x) = 5^(x2-3). The elegance of this form lies in the fact that if two exponential expressions with the same base are equal, then their exponents must also be equal. This principle allows us to take the crucial step of equating the exponents, effectively transitioning from an exponential equation to a more familiar algebraic form. By equating the exponents, we obtain the equation 2x = x^2 - 3. This is a significant breakthrough, as we've now arrived at a quadratic equation, a type of equation that we have well-established methods for solving. The quadratic nature of this equation stems from the presence of the x^2 term. Quadratic equations are ubiquitous in mathematics and physics, appearing in a wide range of applications, from projectile motion to optimization problems. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants. Our equation, 2x = x^2 - 3, can be rearranged into this standard form, setting the stage for us to employ the tools and techniques we've learned for solving quadratic equations. This transition from an exponential equation to a quadratic equation exemplifies the power of mathematical manipulation. By strategically applying the properties of exponents, we've transformed a seemingly complex problem into a more manageable one. The path ahead now involves solving this quadratic equation, which will lead us to the values of x that satisfy the original equation. The journey continues, with each step building upon the previous one, as we steadily approach the solution.

Solving the Quadratic Equation: Unveiling the Roots

Having successfully transformed the original equation 25^x = 5^(x2-3) into the quadratic equation 2x = x^2 - 3, our focus now shifts to solving this algebraic expression. To solve the quadratic equation, the first step is to rearrange it into the standard form ax^2 + bx + c = 0. By subtracting 2x from both sides, we get x^2 - 2x - 3 = 0. Now, we have a quadratic equation in the standard form, where a = 1, b = -2, and c = -3. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is the most straightforward approach. We are looking for two numbers that multiply to -3 and add up to -2. These numbers are -3 and 1. Therefore, we can factor the quadratic equation as (x - 3)(x + 1) = 0. This factored form provides us with a direct path to the solutions. For the product of two factors to be zero, at least one of the factors must be zero. Therefore, either x - 3 = 0 or x + 1 = 0. Solving these linear equations, we find that x = 3 or x = -1. These are the roots of the quadratic equation, the values of x that make the equation true. The process of factoring a quadratic equation is a testament to the power of algebraic manipulation. By recognizing the pattern and applying the appropriate techniques, we can break down a complex expression into simpler components, revealing the underlying solutions. The roots of the quadratic equation are not just abstract numbers; they are the keys to unlocking the solution of the original exponential equation. We have now found two potential values of x that could satisfy the initial equation. The next step is to verify these solutions, ensuring that they indeed make the original equation true.

Verifying the Solutions: Ensuring Accuracy and Validity

Having obtained the potential solutions x = 3 and x = -1 for the equation 25^x = 5^(x2-3), it's crucial to verify these solutions. This verification process is not merely a formality; it's a fundamental step in ensuring the accuracy and validity of our results. In mathematics, it's always prudent to double-check our work, especially when dealing with equations that involve exponents and algebraic manipulations. To verify the solutions, we substitute each value of x back into the original equation and check if both sides of the equation are equal. Let's start with x = 3. Substituting x = 3 into the original equation, we get 25^3 = 5⁽³2-3). Simplifying the left side, we have 25^3 = 15625. On the right side, we have 5⁽³2-3) = 5^(9-3) = 5^6 = 15625. Since both sides are equal, x = 3 is indeed a valid solution. Now, let's verify x = -1. Substituting x = -1 into the original equation, we get 25^(-1) = 5^((-1)2-3). Simplifying the left side, we have 25^(-1) = 1/25. On the right side, we have 5^((-1)2-3) = 5^(1-3) = 5^(-2) = 1/25. Since both sides are equal, x = -1 is also a valid solution. The verification process not only confirms the correctness of our solutions but also reinforces our understanding of the equation and the steps we took to solve it. It's a testament to the rigor and precision that are hallmarks of mathematical thinking. By meticulously checking our work, we can confidently assert that the solutions we've obtained are accurate and reliable. With the solutions verified, we can now confidently state the values of x that satisfy the equation.

The Solutions Unveiled: x = -1 and x = 3

After a rigorous journey through the realms of exponents, algebraic manipulation, and verification, we've arrived at the final destination: the solutions to the equation 25^x = 5^(x2-3). Our exploration began with the recognition of a common base, allowing us to transform the exponential equation into a more manageable quadratic form. We then skillfully navigated the process of solving the quadratic equation, employing the technique of factoring to unveil the potential solutions. And finally, with meticulous attention to detail, we verified these solutions, ensuring their accuracy and validity. The culmination of our efforts reveals that the values of x that satisfy the equation are x = -1 and x = 3. These are the precise values that, when substituted into the original equation, make the statement true. The solutions are not just numbers; they are the answers to the puzzle, the culmination of a logical and systematic process. They represent the points where the exponential functions on either side of the equation intersect, the values where the relationship holds. This journey of solving for x has been more than just a mathematical exercise; it's been a testament to the power of problem-solving, the elegance of mathematical reasoning, and the satisfaction of arriving at a definitive answer. The solutions x = -1 and x = 3 stand as a testament to our perseverance and the effectiveness of the mathematical tools we've employed. As we conclude this exploration, we can appreciate the beauty and precision of mathematics, its ability to unravel complex problems and reveal the underlying truths.

Final Answer: The final answer is $\boxed{x=-1, x=3}$