E 2x 3e X 2 0

Deconstructing and Solving e^(2x) - 3e^x + 2 = 0: A Comprehensive Guide

The equation e^(2x) - 3e^x + 2 = 0 might look intimidating at first glance, but with a systematic approach and understanding of exponential properties, it's entirely solvable. This comprehensive guide will break down the solution step-by-step, exploring the underlying mathematical principles and offering insights into similar problem-solving strategies. We'll cover various methods, ensuring a thorough grasp of this type of exponential equation.

Understanding the Equation: A Foundation in Exponentials

Before diving into the solution, let's analyze the structure of the equation: e^(2x) - 3e^x + 2 = 0. Notice that the equation involves exponential terms with the base e (Euler's number, approximately 2.71828). The key to solving this lies in recognizing that e^(2x) can be rewritten as (e^x)². This transformation is crucial for simplifying the equation and enabling a straightforward solution method.

Method 1: Substitution for Simplification

This approach utilizes a substitution to transform the equation into a quadratic form, making it significantly easier to solve.

1. The Substitution:

Let's introduce a substitution variable, u = e^x. This substitution simplifies the equation considerably. Replacing e^x with u, our equation becomes:

u² - 3u + 2 = 0

2. Solving the Quadratic Equation:

This is now a standard quadratic equation, easily solvable using various methods such as factoring, the quadratic formula, or completing the square. Factoring is the most efficient method in this case:

(u - 1)(u - 2) = 0

This gives us two possible solutions for u:

u = 1 or u = 2

3. Back-Substitution and Solving for x:

Remember that our substitution was u = e^x. Now, we need to substitute back to solve for x:

• For u = 1:

e^x = 1

Taking the natural logarithm (ln) of both sides:

ln(e^x) = ln(1)

x = 0

• For u = 2:

e^x = 2

Taking the natural logarithm of both sides:

ln(e^x) = ln(2)

x = ln(2)

Therefore, the solutions to the original equation e^(2x) - 3e^x + 2 = 0 are x = 0 and x = ln(2).

Method 2: Direct Factoring (Advanced Approach)

While substitution is generally preferred for its clarity, it's possible to factor the original equation directly. This approach requires a more advanced understanding of exponential properties.

The equation e^(2x) - 3e^x + 2 = 0 can be factored as:

(e^x - 1)(e^x - 2) = 0

This factorization directly leads to the same solutions as before:

e^x - 1 = 0 => e^x = 1 => x = 0

e^x - 2 = 0 => e^x = 2 => x = ln(2)

This method, while elegant, can be more challenging to recognize initially. The substitution method is generally recommended for its accessibility.

Verifying the Solutions

It's crucial to verify our solutions by substituting them back into the original equation:

For x = 0:

e^(2*0) - 3e^0 + 2 = 1 - 3 + 2 = 0 (Correct)

For x = ln(2):

e^(2ln(2)) - 3e^(ln(2)) + 2 = e^(ln(2²)) - 3(2) + 2 = 4 - 6 + 2 = 0 (Correct)

Both solutions are confirmed to satisfy the original equation.

Exploring the Graphical Representation

Visualizing the equation can provide further understanding. The equation e^(2x) - 3e^x + 2 = 0 can be represented graphically by plotting the function y = e^(2x) - 3e^x + 2 and observing where it intersects the x-axis (where y = 0). The x-intercepts will correspond to the solutions we've found (x = 0 and x = ln(2)). This graphical representation offers a different perspective on the problem and reinforces the validity of the solutions.

Expanding the Concept: Similar Exponential Equations

The techniques used to solve e^(2x) - 3e^x + 2 = 0 are applicable to a wider range of exponential equations. Any equation that can be manipulated to resemble a quadratic form after an appropriate substitution can be solved using similar methods. For example, equations like:

• 2^(2x) - 7(2^x) + 12 = 0
• 3^(2x) + 2(3^x) - 15 = 0

can be approached with similar substitution strategies, where you'd let u = 2^x or u = 3^x respectively. The key is recognizing the underlying quadratic structure hidden within the exponential terms.

Advanced Considerations and Applications

While we've focused on the core solution methods, this type of equation has wider implications in various fields:

• Differential Equations: Exponential functions frequently appear in solutions to differential equations, particularly those modeling growth or decay processes. Understanding the solution methods for these types of algebraic equations is foundational for working with differential equations.
• Mathematical Modeling: Exponential functions are powerful tools for modeling real-world phenomena, such as population growth, radioactive decay, and compound interest. The ability to solve equations involving exponential terms is vital for applying these models effectively.
• Calculus: This type of equation often arises in calculus problems involving exponential functions, such as optimization problems or the determination of critical points.

Mastering the solution techniques presented here is not merely an exercise in algebra; it's a building block for tackling more complex mathematical problems in various disciplines.

Conclusion: A Powerful Tool for Problem Solving

Solving the equation e^(2x) - 3e^x + 2 = 0 is a valuable exercise that demonstrates the power of substitution and the importance of recognizing hidden quadratic structures. By applying these techniques, you can confidently tackle similar exponential equations and build a stronger foundation in mathematical problem-solving. Remember that verifying your solutions and considering the graphical representation are crucial steps in ensuring accuracy and understanding. This equation serves as a gateway to understanding more complex mathematical concepts and applications.