Solving Exponential Equations Finding X For 25^x = 5^(x^2-3)

In this article, we will delve into solving an exponential equation. Exponential equations, where the variable appears in the exponent, often require us to manipulate the equation using properties of exponents and logarithms to isolate the variable. Here, we aim to find the values of x that satisfy the equation 25^x = 5^(x2-3). This problem combines the concepts of exponential functions and quadratic equations, making it a great exercise for reinforcing fundamental mathematical principles. To successfully tackle this, we'll leverage the common base method, which involves expressing both sides of the equation with the same base, thereby allowing us to equate the exponents. Let’s explore the step-by-step approach to solving this equation and understand the underlying concepts.

The given exponential equation is:

25^x = 5^(x2-3)

Our objective is to determine the values of x that make this equation true. To achieve this, we need to manipulate the equation to a form where we can directly compare the exponents. This involves using properties of exponents and potentially solving a resulting algebraic equation. By finding these values, we gain a deeper understanding of how exponential functions behave and how to solve equations involving them.

1. Express Both Sides with the Same Base

The key to solving this equation is to express both sides with the same base. We know that 25 can be written as 5^2. Therefore, we rewrite the left side of the equation:

25^x = (5²⁾x

Using the power of a power rule, which states that (a^m)n = a^(mn), we get:

(5²⁾x = 5^(2x)

Now, our equation looks like this:

5^(2x) = 5^(x2-3)

By expressing both sides with the same base, we've set the stage for equating the exponents, a crucial step in solving exponential equations.

2. Equate the Exponents

Now that both sides of the equation have the same base (5), we can equate the exponents. This is because if a^m = a^n, then m = n. Applying this principle, we set the exponents equal to each other:

2x = x^2 - 3

This step transforms our exponential equation into a quadratic equation, which is a more familiar form to solve. The transition from exponential to quadratic form is a common technique in solving exponential equations, making it a vital skill for anyone studying algebra and beyond.

3. Rearrange to Form a Quadratic Equation

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

0 = x^2 - 3 - 2x

Rearranging the terms, we get:

x^2 - 2x - 3 = 0

Now we have a quadratic equation in standard form, ready to be factored or solved using the quadratic formula. The ability to manipulate equations into standard forms like this is essential for solving a wide range of mathematical problems.

4. Solve the Quadratic Equation

We can solve the quadratic equation x^2 - 2x - 3 = 0 by factoring. We are looking for two numbers that multiply to -3 and add to -2. These numbers are -3 and 1. Therefore, we can factor the quadratic equation as:

(x - 3)(x + 1) = 0

Setting each factor equal to zero gives us the solutions for x:

x - 3 = 0 or x + 1 = 0

Solving these linear equations, we find:

x = 3 or x = -1

These are the values of x that satisfy our original exponential equation. Factoring is a powerful technique for solving quadratic equations, and it's often the quickest method when the quadratic expression can be easily factored.

5. Verify the Solutions

To ensure our solutions are correct, we should verify them by plugging them back into the original equation:

25^x = 5^(x2-3)

For x = 3:

25^3 = 5⁽³2-3)

25^3 = 5^(9-3)

25^3 = 5^6

(5²⁾3 = 5^6

5^6 = 5^6

This confirms that x = 3 is a solution.

For x = -1:

25^(-1) = 5^((-1)2-3)

25^(-1) = 5^(1-3)

25^(-1) = 5^(-2)

(5²⁾(-1) = 5^(-2)

5^(-2) = 5^(-2)

This confirms that x = -1 is also a solution.

Verification is a crucial step in problem-solving, especially in mathematics, as it ensures the accuracy of our results and helps catch any potential errors in our calculations.

The values of x that satisfy the equation 25^x = 5^(x2-3) are x = -1 and x = 3. These solutions make the exponential equation true, as verified by substituting them back into the original equation. The process involved expressing both sides of the equation with a common base, equating the exponents, forming a quadratic equation, solving for x, and verifying the solutions. This approach is a standard method for solving exponential equations and demonstrates the interconnectedness of different mathematical concepts, such as exponents, quadratics, and equation-solving techniques. Therefore, the correct answer is:

B. x = -1, x = 3

In summary, we have successfully solved the exponential equation 25^x = 5^(x2-3) by applying key principles of exponential functions and quadratic equations. The process involved transforming the equation into a solvable form by expressing both sides with the same base, equating the exponents, and then solving the resulting quadratic equation. We found that the values x = -1 and x = 3 satisfy the original equation. This exercise highlights the importance of understanding the properties of exponents and how they can be used in conjunction with other algebraic techniques to solve complex problems. Mastering these skills is essential for anyone looking to excel in mathematics and related fields. The ability to manipulate and solve exponential equations is a valuable asset in various areas of science, engineering, and finance, where exponential growth and decay models are frequently used. Through practice and a solid understanding of the underlying concepts, one can confidently tackle a wide range of mathematical challenges.