Problem 81 - A Graphical Approach to College Algebra

Question

$$\text { The given equations are quadratic in form. Solve each and give exact solutions.}$$ $$e^{2 x}-6 e^{x}+8=0$$

Step-by-Step Solution

Answer

The exact solutions are \( x = \ln(2) \) and \( x = \ln(4) \).

Step 1: Identify the substitution

Recognize that the equation \( e^{2x} - 6e^x + 8 = 0 \) can be rewritten in quadratic form. To do this, set \( y = e^x \). Then \( e^{2x} = (e^x)^2 = y^2 \). This gives us a new equation: \( y^2 - 6y + 8 = 0 \).

Step 2: Solve the quadratic equation

The equation \( y^2 - 6y + 8 = 0 \) is a standard quadratic equation in terms of \( y \). Factor the equation to find its roots. The factors of 8 that sum up to -6 are -2 and -4. Thus, the equation factors as \((y - 2)(y - 4) = 0\).

Step 3: Find the roots of the quadratic equation

Set each factor equal to zero to find the values of \( y \). This yields: \( y - 2 = 0 \) or \( y - 4 = 0 \). Solving these gives \( y = 2 \) and \( y = 4 \).

Step 4: Back-substitute to find \( x \)

Recall the substitution \( y = e^x \). Replace \( y \) with \( e^x \) in the solutions. So, we have \( e^x = 2 \) and \( e^x = 4 \).

Step 5: Solve for \( x \)

To find \( x \), take the natural logarithm of both sides of the equations. Thus, solving \( e^x = 2 \) gives \( x = \ln(2) \) and solving \( e^x = 4 \) gives \( x = \ln(4) \).

Key Concepts

Exponential Functions

Exponential functions are a type of function where a constant base is raised to a variable exponent. In mathematical notation, they take the form \( a^x \), where \( a \) is the base and \( x \) is the exponent. Exponential functions are essential in modeling growth and decay processes, such as population growth and radioactive decay.
Some key properties of exponential functions include:

• They are always positive for any real number exponent, assuming the base is positive and not equal to zero.
• If the base \( a > 1 \), the function is increasing or grows as \( x \) increases.
• If the base \( 0 < a < 1 \), the function decreases or decays as \( x \) increases.

The special base \( e \), approximately equal to 2.71828, is frequently used in exponential functions, especially in calculus and complex functions. Base \( e \) is natural because it simplifies the process of differentiation and integration. In the quadratic equation \( e^{2x} - 6e^x + 8 = 0 \), \( e^x \) is an exponential function with base \( e \). Understanding how these functions work conceptually and mathematically is crucial for solving various types of if they appear in different contexts.

Natural Logarithms

Natural logarithms are logarithms with base \( e \). Represented as \( \ln(x) \), they are significantly useful in calculus and solving exponential equations. This is because they turn multiplication into addition and exponentiation into multiplication, simplifying many calculus operations.
A natural logarithm answers the question: "To what power must \( e \) be raised to produce \( x \)?" For instance, if \( e^x = y \), then \( x = \ln(y) \). This transformation is particularly helpful in solving our quadratic-like equation, where exponential terms are involved.

• \( \ln(e) = 1 \): Because \( e^1 = e \).
• \( \ln(1) = 0 \): Since \( e^0 = 1 \).
• The function is always increasing, which means it is a one-to-one function, important for ensuring unique solutions.

Using natural logarithms in equations enables us to "undo" exponentiation and find the value of the exponent, like we did when solving for \( x \) in \( e^x = 2 \) or \( e^x = 4 \), leading to solutions \( x = \ln(2) \) and \( x = \ln(4) \). Understanding how to apply natural logarithms is vital when dealing with exponential growth models as well.

Factorization

Factorization is an algebraic technique used to express a polynomial as a product of its simpler, constituent polynomials. This technique simplifies solving polynomial equations, especially quadratics, by breaking them into factors that can easily be solved to find the roots or zeros of the original polynomial.
The process involves identifying values that, when multiplied, yield the constant term and added together, give the middle coefficient in the polynomial. For example, to factorize \( y^2 - 6y + 8 \), we identified -2 and -4 since their product is 8 and their sum is -6. Therefore, the expression factorized to \((y - 2)(y - 4) = 0\).

• Factoring helps find solutions when polynomial equations are set to zero.
• Each factor represents a potential solution, and setting each to zero yields the solution set for the variable, as in our example: \( y = 2 \) and \( y = 4 \).
• It checks if an expression can be expressed even further into simpler parts, enabling more straightforward examination and manipulation.

Mastering factorization is crucial for efficiently tackling quadratic equations, such as transforming complex possibilities into more manageable mathematical problems.