Problem 34 - Algebra and Trigonometry

Question

Solve the equation. $$ e^{x}-12 e^{-x}-1=0 $$

Step-by-Step Solution

Answer

The solution is \( x = \ln(4) \).

Step 1: Substitution

To simplify the equation, first let \( y = e^{x} \). This gives us the equation \( y - \frac{12}{y} - 1 = 0 \).

Step 2: Eliminate the Fraction

Multiply the entire equation by \( y \) to eliminate the fraction:\[ y^2 - 12 - y = 0 \] which simplifies to \( y^2 - y - 12 = 0 \).

Step 3: Solve the Quadratic Equation

The equation \( y^2 - y - 12 = 0 \) is a quadratic equation, which can be solved using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1 \), \( b = -1 \), and \( c = -12 \).

Step 4: Apply the Quadratic Formula

Substitute \( a \), \( b \), and \( c \) into the quadratic formula:\[ y = \frac{1 \pm \sqrt{1 + 48}}{2} \]This simplifies to:\[ y = \frac{1 \pm \sqrt{49}}{2} \]\[ y = \frac{1 \pm 7}{2} \]

Step 5: Calculate the Solutions for y

Calculate the potential values of \( y \):1. \( y = \frac{1 + 7}{2} = 4 \)2. \( y = \frac{1 - 7}{2} = -3 \)Since \( y = e^{x} \) must be positive, \( y = 4 \) is the only valid solution.

Step 6: Solve for x

Recall that \( y = e^{x} \) and we have found \( y = 4 \). Thus: \[ e^{x} = 4 \]Taking the natural logarithm of both sides gives:\[ x = \ln(4) \]

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra. They appear in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, with \( a \) not equal to zero. The equation has up to two solutions, which can be found using various methods such as factoring, completing the square, or the quadratic formula.
The quadratic formula is a universal tool for solving any quadratic equation:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our original problem, after substitution and simplification, we reached the quadratic form \( y^2 - y - 12 = 0 \). Plugging \( a = 1 \), \( b = -1 \), and \( c = -12 \) into the quadratic formula allowed us to find the potential solutions for \( y \).
Understanding these solutions involves calculating the discriminant \( b^2 - 4ac \), which helps determine the nature of the roots. A positive discriminant signifies two distinct real roots, a discriminant of zero marks a perfect square with one real root, and a negative discriminant implies complex roots.

Natural Logarithms

Natural logarithms are a specific type of logarithm with the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. The natural logarithm of a number \( x \) is expressed as \( \ln(x) \), and it helps to reverse the exponential operation, particularly useful in solving equations involving \( e^{x} \).
In the context of the provided problem, once we have \( e^{x} = 4 \), we employ the natural logarithm to isolate \( x \). Therefore, taking the natural logarithm of both sides gives:

• \( x = \ln(4) \)

This step is crucial because it translates an exponential equation into a simple arithmetic expression, making it easier to solve for \( x \). The properties of logarithms, such as the power rule and product identities, are pivotal in algebraic manipulations and simplifications.

Substitution Method

The substitution method is an efficient way to simplify equations, especially when they involve complex expressions like exponentials. It involves replacing a complicated part of an equation with a simpler variable. This is particularly helpful when transforming an equation into a more recognizable form, such as a quadratic equation.
In our example, we utilized substitution by letting \( y = e^{x} \). This conversion reformed the original equation into a quadratic equation, \( y^2 - y - 12 = 0 \), making it significantly easier to solve.

• Step 1: Identify the expression to substitute (e.g., \( e^{x} \)).
• Step 2: Replace it with a simpler variable (\( y \)).
• Step 3: Solve the transformed equation.
• Step 4: Substitute back the original terms to find the final solution.

By simplifying originally complex equations into more palatable forms, substitution becomes a powerful tool in solving exponential and other types of equations efficiently.