Problem 81 - A Graphical Approach to Precalculus with Limits

Question

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

\( x = \ln 4 \) and \( x = \ln 2 \).

Step 1: Identify Substitution for Simplification

The given equation is \( e^{2x} - 6e^x + 8 = 0 \). Notice that this equation resembles a quadratic form. Introduce the substitution \( u = e^x \). Hence, \( e^{2x} = (e^x)^2 = u^2 \). The equation becomes \( u^2 - 6u + 8 = 0 \).

Step 2: Solve the Quadratic Equation

Now, you need to solve the resulting quadratic equation \( u^2 - 6u + 8 = 0 \). This can be solved using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = 8 \). Calculate the discriminant: \( b^2 - 4ac = 36 - 32 = 4 \).

Step 3: Calculate the Roots Using the Quadratic Formula

Substitute the values into the quadratic formula:\[ u = \frac{-(-6) \pm \sqrt{4}}{2(1)} = \frac{6 \pm 2}{2} \]This provides the solutions \( u = \frac{8}{2} = 4 \) and \( u = \frac{4}{2} = 2 \). Thus, \( u = 4 \) or \( u = 2 \).

Step 4: Back-substitute to Original Variable

Recall the substitution \( u = e^x \). So the equations become \( e^x = 4 \) and \( e^x = 2 \). Taking the natural logarithm of both sides of these equations gives \( x = \ln 4 \) and \( x = \ln 2 \).

Step 5: Write the Exact Solutions

The solutions to the original problem are \( x = \ln 4 \) and \( x = \ln 2 \). These are the exact values based on the properties of logarithms.

Key Concepts

Exponential Equations

Exponential equations are equations where variables appear in the exponent. For example, in the equation \( e^{2x} - 6e^x + 8 = 0 \), the variable \( x \) is located in the exponent position. Solving exponential equations often requires specific strategies, such as the substitution method, to simplify the process. These equations become quite manageable once you transform them into more familiar forms like quadratic equations.

• First, look for patterns that might suggest substitution possibilities.
• Transform the exponential terms into a simpler variable, such as setting \( u = e^x \).
• After substitution, solve the resulting equation.

Transforming exponential equations into different forms is key to solving them accurately and efficiently.

Substitution Method

The substitution method is a technique used to simplify complex equations by introducing a new variable. This new variable makes the equation resemble a more familiar form. In the given example, the original exponential equation \( e^{2x} - 6e^x + 8 = 0 \) is made easier to solve by using the substitution \( u = e^x \). This transforms the equation into a quadratic form \( u^2 - 6u + 8 = 0 \).

• Choose an appropriate substitution to simplify the equation.
• Ensure that the substitution makes the equation easier to solve, possibly transforming it into a linear or quadratic equation.
• Once solved, substitute back the original terms to find the actual solution of the problem.

Using the substitution method not only simplifies problems but also helps find the solution efficiently.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our example, once the substitution \( u = e^x \) was made, we ended up with the quadratic equation \( u^2 - 6u + 8 = 0 \).

Here are the steps involved in applying the quadratic formula:

• Identify the coefficients: \( a = 1 \), \( b = -6 \), \( c = 8 \).
• Calculate the discriminant: \( b^2 - 4ac = 36 - 32 = 4 \).
• Apply the formula to find the roots: \( u = \frac{-(-6) \pm \sqrt{4}}{2(1)} = \frac{6 \pm 2}{2} \), yielding \( u = 4 \) and \( u = 2 \).

The quadratic formula is a powerful method to find the roots of any quadratic equation, and when integrated with substitution methods, it solves a variety of complex problems efficiently.