Problem 33 - Algebra and Trigonometry

Question

Solve the equation. $$ e^{4 x}+4 e^{2 x}-21=0 $$

Step-by-Step Solution

Answer

\(x = \frac{1}{2}\ln(3)\).

Step 1: Substitution

To simplify the equation, let us substitute a new variable. Let \(y = e^{2x}\). This implies that \(y^2 = (e^{2x})^2 = e^{4x}\). So, the equation becomes \(y^2 + 4y - 21 = 0\).

Step 2: Quadratic Equation

We now need to solve the quadratic equation \(y^2 + 4y - 21 = 0\). To do this, we will use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = -21\).

Step 3: Calculate Discriminant

First, calculate the discriminant \(b^2 - 4ac\): \(4^2 - 4 \times 1 \times (-21) = 16 + 84 = 100\).

Step 4: Calculate Roots

Using the quadratic formula, calculate the roots: \(y = \frac{-4 \pm \sqrt{100}}{2 \times 1}\). This gives \(y = \frac{-4 \pm 10}{2}\).

Step 5: Simplify the Roots

Calculate the two solutions: \(y_1 = \frac{-4 + 10}{2} = 3\) and \(y_2 = \frac{-4 - 10}{2} = -7\).

Step 6: Solve for \(x\)

Convert back to \(x\) using the relation \(y = e^{2x}\). For \(y_1 = 3\), solve \(e^{2x} = 3\), which gives \(2x = \ln(3)\), thus \(x = \frac{1}{2}\ln(3)\). For \(y_2 = -7\), there is no solution because \(e^{2x}\) cannot be negative.

Step 7: Final Solution

The valid solution for the equation based on realistic values is \(x = \frac{1}{2}\ln(3)\).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra, representing polynomial equations of degree two. The standard form of a quadratic equation is written as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are coefficients, and \(x\) represents the variable. Quadratics are characterized by their distinctive parabolic graphs.Some key features include:

• The highest exponent of the variable is 2.
• The graph of a quadratic function is always a parabola.
• Quadratics can have zero, one, or two real solutions.

In our exercise, we transformed the equation into a quadratic form using substitution, which allowed us to leverage familiar techniques to find the solution.

Substitution Method

The substitution method involves replacing a complex component of an equation with a simpler variable to make calculations more straightforward. It is particularly useful when dealing with exponential equations that can be rewritten as polynomial equations.For the exercise given, the substitution \(y = e^{2x}\) transformed the original exponential equation \(e^{4x} + 4e^{2x} - 21 = 0\) into the quadratic equation \(y^2 + 4y - 21 = 0\).This step:

• Simplifies the equation by reducing the complexity of the exponentials.
• Makes it easier to apply mathematical techniques such as the quadratic formula.

After solving for \(y\), we can revert back to the original variables to solve for \(x\). This technique streamlines the process and helps in focusing on the solution of quadratic equations.

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions of any quadratic equation\(ax^2 + bx + c = 0\). The formula is given by:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Using this formula requires calculating the discriminant \(b^2 - 4ac\) first. The discriminant helps determine the nature of the roots:

• If it is positive, there are two real solutions.
• If it is zero, there is one real solution.
• If negative, there are no real solutions.

In our problem, with \(a = 1\), \(b = 4\), and \(c = -21\), the discriminant was calculated to be 100, indicating two real solutions for \(y\). The solutions were then found using the formula, and it was determined that only one value of \(y\) leads to a valid value for \(x\).

Exponential Functions

Exponential functions are characterized by a variable exponent, typically represented as \(f(x) = a^x\), where \(a\) is a constant and \(x\) is the variable. These functions are crucial in modeling growth and decay processes in real life.In the context of solved equations, the exponential components \(e^{4x}\) and \(e^{2x}\) needed simplifying, which we achieved through substitution. The nature of exponentials, such as \(e^{2x}\), is essential to understanding why only non-negative outputs are physically manageable, as indicated with \(e^{2x} = y\). Negative values for \(y\) imply no real solutions for \(x\), since exponentials never yield negative outcomes.Applying our understanding of exponentials, we solved for \(x\) by transforming back from the variable \(y\), highlighting the versatility of exponential functions in algebraic solutions.