Problem 32 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 25-32: Solve the equation without using a calculator. $$ e^{x}+4 e^{-x}=5 $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = \ln(4) \).

Step 1: Identify the Equation and Introduce a Substitution

The given equation is \( e^x + 4e^{-x} = 5 \). To simplify, introduce the substitution \( y = e^x \). This implies that \( e^{-x} = \frac{1}{y} \). Substitute this into the equation to get \( y + \frac{4}{y} = 5 \).

Step 2: Clear the Fraction

Multiply both sides of the equation by \( y \) to eliminate the fraction: \( y^2 + 4 = 5y \).

Step 3: Rearrange into a Quadratic Equation

Rearrange the equation into standard quadratic form: \( y^2 - 5y + 4 = 0 \).

Step 4: Solve the Quadratic Equation

Factor the quadratic equation: \( (y - 1)(y - 4) = 0 \). From this, we find the solutions \( y = 1 \) and \( y = 4 \).

Step 5: Revert the Substitution

Recall that \( y = e^x \). Therefore, from \( y = 1 \), we get \( e^x = 1 \), so \( x = \ln(1) = 0 \). From \( y = 4 \), we get \( e^x = 4 \), so \( x = \ln(4) \).

Step 6: Write the Final Solutions

The solutions to the original equation are \( x = 0 \) and \( x = \ln(4) \).

Key Concepts

Substitution Method

Solving complex equations often becomes simpler with the Substitution Method. It involves replacing part of an expression with a new variable. In our exercise, we have the equation \( e^x + 4e^{-x} = 5 \). Instead of dealing with both \( e^x \) and \( e^{-x} \), we introduce a substitution: let \( y = e^x \).
This means \( e^{-x} \) can be rewritten as \( \frac{1}{y} \), simplifying our equation to \( y + \frac{4}{y} = 5 \).
This substitution transforms a seemingly complicated exponential equation into a more familiar form, making the problem easier to solve.

**Why use Substitution?**

• Simplifies complex expressions.
• Helps in transforming the problem into familiar territory (e.g., polynomial form).
• Makes calculations easier, especially when dealing with inversely related terms.

Quadratic Equations

Once we've introduced the substitution \( y = e^x \), the equation \( y + \frac{4}{y} = 5 \) can be rewritten by clearing the fraction. By multiplying everything by \( y \), we get \( y^2 + 4 = 5y \).
Rearranging this gives us \( y^2 - 5y + 4 = 0 \), which is a quadratic equation.
Quadratic equations, expressed in the form \( ax^2 + bx + c = 0 \), are a type of polynomial equation.

**Characteristics of Quadratic Equations:**

• They have at most two solutions.
• The solutions can be found using factoring, completing the square, or the quadratic formula.
• The solutions can be real or complex numbers depending on the discriminant \( b^2 - 4ac \).

In this scenario, the discriminant \( (-5)^2 - 4 \times 1 \times 4 = 9 \) is positive, indicating two distinct real solutions.

Factoring

Once the quadratic equation \( y^2 - 5y + 4 = 0 \) is in standard form, we can use factoring to find the solutions.
Factoring involves expressing the quadratic as a product of two binomials. For this equation, that means finding two numbers that multiply to 4 (the constant term) and add up to -5 (the coefficient of \(y\)).
The factors \( (y - 1)(y - 4) = 0 \) satisfy these conditions.

**Steps for Factoring Quadratic Equations:**

• Identify two numbers that multiply to \( c \) and add to \( b \).
• Write the expression as a product of binomials: \( (y - 1)(y - 4) = 0 \).
• Set each factor equal to zero: \( y - 1 = 0 \) or \( y - 4 = 0 \).
• Solve for \( y \) to get the roots of the quadratic equation.

In this case, the solutions are \( y = 1 \) and \( y = 4 \).

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are used to solve equations involving \( e \), the base of natural logarithms. After solving for \( y \) using substitution, we revert to the original variable \( x \).
Given \( y = e^x \), for \( y = 1 \), we have \( e^x = 1 \), which implies \( x = \ln(1) = 0 \).
For \( y = 4 \), we have \( e^x = 4 \), leading to \( x = \ln(4) \).

**Understanding Natural Logarithms:**

• \( \ln(a) \) gives the power to which \( e \) must be raised to equal \( a \).
• \( \ln(1) = 0 \) because \( e^0 = 1 \).
• Useful for solving equations involving exponentials, making them linear.

The final solutions to the original equation \( e^x + 4e^{-x} = 5 \) are therefore \( x = 0 \) and \( x = \ln(4) \).