Problem 49 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-8 e^{x}+12=0\)

Step-by-Step Solution

Answer

The solution to the equation \(e^{2 x}-8 e^{x}+12=0\) are \(x = 0.693\) and \(x = 1.792\) after approximating to three decimal places.

Step 1: Transform the Equation

Transform the original exponential equation \(e^{2 x}-8 e^{x}+12=0\) into a more familiar form, a quadratic equation. This can be done by letting \(y=e^{x}\), so the equation becomes \(y^{2} - 8y + 12 = 0\).

Step 2: Factor the Quadratic Equation

Factor the quadratic equation \(y^{2} - 8y + 12 = 0\) . This equation can be factored into \((y-2)(y-6)=0\).

Step 3: Apply Zero Product Property

By the zero product property, if the product of two factors equals zero then at least one of the factors must be equal to zero. So, we set each factor equal to zero and solve for \(y\), yielding solutions: \(y = 2\) and \(y = 6\). But remember, \(y = e^{x}\).

Step 4: Find the solution for x

Substitute \(e^{x}\) back in place of \(y\) and solve for \(x\). So, \(e^{x} = 2\) and \(e^{x} = 6\). Solving for \(x\), we get \(x = ln(2)\) and \(x = ln(6)\).

Step 5: Approximate the results

Using a calculator to find the natural logarithm of 2 and 6, we get our solutions to three decimal places which are \(x = 0.693\) and \(x = 1.792\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental to algebra and appear in various forms throughout mathematics and science. These equations are of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). The solutions to these equations, also known as the roots, can be found using several methods, such as factoring, completing the square, or applying the quadratic formula. Factoring is often the quickest method when it is feasible, as in the exercise where \(e^{2x} - 8e^{x} + 12 = 0\) is transformed into a factored form through substitution. Understanding how to recognize and solve quadratic equations is crucial for tackling a wide array of problems in algebra and beyond.

They are often applicable to real-world scenarios, such as projectile motion and optimizing area and volume problems. In the context of solving exponential equations algebraically, recognizing that an equation can be rewritten in quadratic form provides a powerful strategy to find solutions.

Zero Product Property

The zero product property is an essential algebraic principle that states if the product of two factors equals zero, then at least one of the factors must be zero. Formally, if \(a \cdot b = 0\), then either \(a = 0\) or \(b = 0\) (or both). This property is particularly handy when solving quadratic equations after factoring. By setting each factor equal to zero, you can solve for the variable of interest. In our example, after factoring \(e^{2x} - 8e^{x} + 12 = 0\) into \( (e^{x} - 2)(e^{x} - 6) = 0 \) , the zero product property allows us to deduce that \(e^{x} - 2 = 0\) or \(e^{x} - 6 = 0\). This property simplifies the process of finding the exact values of \(x\) that satisfy the original exponential equation. The facility with the zero product property is therefore critical for solving not only quadratic equations but also various polynomial equations.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is the mathematical constant approximately equal to 2.71828. This special logarithm is integral to solving equations where the variable is an exponent, as is the case with exponential equations. When solving for \(x\) in equations such as \(e^{x} = y\), taking the natural logarithm of both sides allows us to extract the exponent: \(x = \ln(y)\).

In our exercise, we apply this step after utilizing the zero product property to find \(e^{x} = 2\) and \(e^{x} = 6\), leading to \(x = \ln(2)\) and \(x = \ln(6)\). The natural logarithm thus provides a bridge between exponential expressions and their corresponding exponents, enabling us to solve for \(x\) and approximate its value using a calculator. Appreciating the relationship between the exponential function and the natural logarithm is pivotal for students to gain proficiency in algebra and to understand growth and decay processes in advanced mathematics and applied sciences.