Problem 82 - A Graphical Approach to Precalculus with Limits

Question

The given equations are quadratic in form. Solve each and give exact solutions. $$e^{2 x}-8 e^{x}+15=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \ln 3 \) and \( x = \ln 5 \).

Step 1: Identify the Substitution

The given equation is \( e^{2x} - 8e^x + 15 = 0 \). Notice the expression is quadratic in form with substitution. Let \( y = e^x \). This implies \( y^2 = e^{2x} \). Thus, rewrite the equation as \( y^2 - 8y + 15 = 0 \).

Step 2: Solve the Quadratic Equation

The quadratic equation \( y^2 - 8y + 15 = 0 \) needs to be solved. It can be factored into \( (y - 3)(y - 5) = 0 \). Thus, the solutions for \( y \) are \( y = 3 \) and \( y = 5 \).

Step 3: Back Substitute to Solve for \( x \)

Recall the substitution made: \( y = e^x \). Substituting back we have two equations: \( e^x = 3 \) and \( e^x = 5 \). Solve for \( x \) by taking the natural logarithm, \( x = \ln 3 \) or \( x = \ln 5 \).

Step 4: State the Exact Solutions

The solutions for the equation \( e^{2x} - 8e^x + 15 = 0 \) are \( x = \ln 3 \) and \( x = \ln 5 \). These are the exact solutions in logarithmic form.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In the context of this exercise, the function involves the base of the natural logarithm, which is Euler's number, denoted as \( e \).
This type of function has the form \( e^x \), where the base \( e \) is approximately equal to 2.71828.
The power \( x \) is a variable which can represent any real number.

• The nature of exponential functions is that they grow rapidly as the exponent increases.
• They are commonly used in contexts where growth processes are modeled, such as population growth or radioactive decay.
• In our exercise, \( e^x \) and \( e^{2x} \) are both examples of exponential terms that are part of a quadratic equation in "quadratic form."

Exponential functions have interesting properties such as the fact that the derivative of \( e^x \) with respect to \( x \) is itself \( e^x \). This self-replicating property in differentiation is unique to the exponential function with base \( e \).

Substitution Method

The substitution method is a valuable technique when solving equations that are initially difficult to handle in their original form. It involves replacing a complex part of an equation with a simpler variable, often simplifying the process of finding a solution.
In this exercise, we used substitution to transform the equation into a more manageable format.For the given equation \( e^{2x} - 8e^x + 15 = 0 \), substitution involves:

• Recognizing that \( e^{2x} \) can be expressed as \( (e^x)^2 \), prompting us to make the substitution \( y = e^x \).
• This changes the equation into a familiar quadratic form: \( y^2 - 8y + 15 = 0 \).

Substitution simplifies solving quadratic equations by reducing exponential components into terms that are easier to handle.
Once solved, we return to the original variable by reversing the substitution, allowing us to find solutions like \( x = \ln 3 \) and \( x = \ln 5 \). This approach exemplifies how substitution can streamline complex calculations.

Factoring

Factoring is a fundamental method used to solve quadratic equations. For an equation in the standard form \( ax^2 + bx + c = 0 \), factoring involves expressing it as a product of two binomial expressions.
This process can help identify the solutions or roots of the equation rapidly. In the example from our exercise, once the substitution was made, the equation \( y^2 - 8y + 15 = 0 \) was solved by factoring:

• Notice the quadratic expression can be rewritten as \( (y - 3)(y - 5) = 0 \).
• The roots of the equation, or solutions for \( y \), are determined by setting each factor to zero: \( y - 3 = 0 \) leading to \( y = 3 \), and \( y - 5 = 0 \) leading to \( y = 5 \).

Factoring is preferred when the quadratic can be expressed neatly in this format, without the need for more complex methods such as completing the square or using the quadratic formula.
It’s a quick and intuitive process, allowing for the conversion of quadratic expressions into their factors, revealing possible solutions efficiently.