Problem 29 - Algebra and Trigonometry

Question

Solve the equation. \(e^{2 x}-3 e^{x}+2=0\)

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = \ln(2) \).

Step 1: Identify the type of equation

The equation given is a quadratic equation in disguise. By letting the substitution \( u = e^x \), the equation becomes \( u^2 - 3u + 2 = 0 \).

Step 2: Substitute and simplify

Once you substitute \( u = e^x \), the equation transforms to \( u^2 - 3u + 2 = 0 \). This is a standard quadratic equation that can be solved using various methods for quadratics.

Step 3: Solve the quadratic equation

Factor the quadratic equation \( u^2 - 3u + 2 = 0 \) into \((u - 1)(u - 2) = 0 \). Therefore, the solutions for \( u \) are \( u = 1 \) and \( u = 2 \).

Step 4: Solve for original variable \(x\)

Since \( u = e^x \), we substitute back to find \( e^x = 1 \) and \( e^x = 2 \). For \( e^x = 1 \), we get \( x = \ln(1) = 0 \). For \( e^x = 2 \), we get \( x = \ln(2) \).

Step 5: Summary of Solution

The solutions for the equation \( e^{2x} - 3e^x + 2 = 0 \) are \( x = 0 \) and \( x = \ln(2) \).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra, characterized by their general form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). They exhibit a parabolic graph and can have one, two, or no real solutions, depending on the discriminant \( b^2 - 4ac \). In the context of our problem, it appears initially as an exponential form, but through substitution, it translates into a quadratic form. This quadratic representation facilitates straightforward solutions, highlighting the versatility of quadratic equations in solving diverse algebraic problems. When the discriminant is positive, as in this exercise where it equates to 1, it suggests two distinct real solutions, confirming that our equation can be factored neatly.

Substitution Method

The substitution method is a valuable algebraic tool often used to simplify complex equations by replacing variables with simpler expressions. In this exercise, substituting \( u = e^x \) transforms the complex exponential equation into a quadratic one. This method breaks down difficult expressions into more manageable forms, enabling easier solution paths. By substituting \( e^x \) with \( u \), the original equation \( e^{2x} - 3e^x + 2 = 0 \) simplifies to \( u^2 - 3u + 2 = 0 \).

• By turning a non-linear problem into a linear one, substitution can directly reveal the problem's structure, simplifying the solving process substantially.
• This approach is especially effective when dealing with equations involving exponential, logarithmic, or trigonometric functions.

It's crucial to revert the solution back to the original variable after solving the simplified equation, ensuring the final solution addresses the original problem.

Factoring Quadratics

Factoring quadratics is a common method used to find the roots of a quadratic equation. The process involves expressing the quadratic as a product of two binomials. In this exercise, once the substitution led to the equation \( u^2 - 3u + 2 = 0 \), factoring the quadratic became the next logical step. We identified its factors as \((u - 1)(u - 2) = 0\), leading to the solutions \( u = 1 \) and \( u = 2 \).

• Factoring relies on identifying two numbers that multiply to give \( c \) and add to give \( b \), a straightforward task when the quadratic is uncomplicated.
• It provides a clean and efficient way to find the zeros of a polynomial equation.

Once factored, set each factor to zero and solve for the variable. This technique highlights the importance of recognizing patterns in algebra and leveraging them for solving equations efficiently.

Natural Logarithms

Natural logarithms are based on the constant \( e \), approximately equal to 2.71828, and are denoted as \( \ln \). They are the inverse functions of the exponential function. In this exercise, we used natural logarithms to find \( x \) once the equations \( e^x = 1 \) and \( e^x = 2 \) were obtained. Applying the property \( \ln(e^x) = x \), it was straightforward to compute:

• \( x = \ln(1) = 0 \), because the natural log of 1 is always 0.
• \( x = \ln(2) \), providing a more precise value than a decimal approximation.

Natural logarithms are vital in solving exponential equations since they allow us to convert the exponential function into a linear one. This bridging method is crucial as it can reduce the complexity of exponential expressions, allowing for simple arithmetic operations to yield the solution.