Problem 32 - Algebra and Trigonometry

Question

Solve the equation. $$ e^{2 x}-e^{x}-6=0 $$

Step-by-Step Solution

Answer

The solution is \( x = \ln(3) \).

Step 1: Recognize the Form of the Equation

The equation given is an exponential equation: \( e^{2x} - e^x - 6 = 0 \). Notice that \( e^{2x} \) can be rewritten as \((e^x)^2 \). This suggests that a substitution can help simplify the equation.

Step 2: Make a Substitution

Let \( y = e^x \). Then \( e^{2x} = y^2 \). Substitute these into the equation to transform it into a quadratic equation: \( y^2 - y - 6 = 0 \).

Step 3: Solve the Quadratic Equation

The transformed equation is \( y^2 - y - 6 = 0 \). Factor the quadratic: \( (y - 3)(y + 2) = 0 \). This gives us the solutions \( y = 3 \) and \( y = -2 \).

Step 4: Back-Substitute to Find x

Recall that \( y = e^x \). For \( y = 3 \), we have \( e^x = 3 \). Take the natural logarithm to solve for \( x \): \( x = \ln(3) \). For \( y = -2 \), since the exponential function \( e^x \) is always positive, there is no solution for \( y = -2 \).

Step 5: Verify the Solutions

Substitute \( x = \ln(3) \) back into the original equation to verify it satisfies the equation: \( e^{2(\ln(3))} - e^{\ln(3)} - 6 = 9 - 3 - 6 = 0 \). It holds true, confirming \( x = \ln(3) \) as a solution. The solution \( y = -2 \) gives no valid exponent cause, confirming only \( x = \ln(3) \) is correct.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree two. It generally takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. The solutions of a quadratic equation are often referred to as "roots." Quadratic equations can be solved by several methods, such as:

• Factoring: Expressing the equation as a product of its factors, if possible.
• Quadratic Formula: Using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions.
• Completing the Square: Rewriting the equation in a perfect square trinomial form.

In the context of the exercise, we transformed an exponential equation into a quadratic equation \( y^2 - y - 6 = 0 \) by substituting \( y = e^x \). Factoring provided the solutions \( y = 3 \) and \( y = -2 \). These solutions were then used to find the corresponding values of \( x \). Quadratic equations are versatile and appear in many mathematical contexts, making them essential to understand thoroughly.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with a base of the mathematical constant \( e \), approximately equal to 2.71828. It is a fundamental concept in mathematics, especially in calculus and solving exponential equations. A natural logarithm \( \ln(a) \) is defined as the power to which \( e \) must be raised to get the number \( a \). Key properties of natural logarithms include:

• \( \ln(1) = 0 \): because \( e^0 = 1 \).
• \( \ln(e) = 1 \): because \( e^1 = e \).
• \( \ln(a \cdot b) = \ln(a) + \ln(b) \): logarithm of a product is the sum of logarithms.
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \): logarithm of a quotient is the difference of logarithms.
• \( \ln(a^b) = b \cdot \ln(a) \): logarithm of a power is the exponent times the logarithm.

In solving our exercise, the natural logarithm was used to find \( x \) from \( e^x = 3 \). This led to \( x = \ln(3) \), providing a crucial step to solving the initial equation.

Substitution Method

The substitution method is a problem-solving technique often used to simplify complex equations, particularly when dealing with exponential or trigonometric forms. This method involves introducing a new variable to replace a more complex expression, making the equation easier to handle.In our original exercise, the substitution method was used to transform an exponential equation into a quadratic form. The steps were:

• Identify a complex component of the equation that looks quadratic-like.
• Set a new variable, say \( y \), equal to this complex component; for instance, \( y = e^x \).
• Express other parts of the equation in terms of this new variable.
• Solve the transformed equation using standard methods like factoring or the quadratic formula.
• Substitute back to find the original variables.

Using substitution in our exercise, we replaced \( e^x \) with \( y \), making the equation \( y^2 - y - 6 = 0 \). After solving this quadratic equation for \( y \), we then returned to the original variable \( x \) using the relationship \( y = e^x \). Substitution simplifies equations and provides clarity, making it a handy tool in mathematics.