Problem 83 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The solution is \(x = \ln\left(\frac{3}{2}\right)\).

Step 1: Observe the Equation

The given equation is \(2e^{2x} + e^{x} = 6\). Notice that this resembles a quadratic equation in terms of \(e^x\). We aim to rewrite it in that form.

Step 2: Make a Substitution

Let \(u = e^x\). The equation \(2e^{2x} + e^{x} = 6\) can be rewritten as \(2u^2 + u = 6\).

Step 3: Rearrange into Quadratic Form

Rearrange the equation in standard quadratic form: \(2u^2 + u - 6 = 0\).

Step 4: Solve the Quadratic Equation

Now solve the quadratic equation \(2u^2 + u - 6 = 0\) using either factoring, the quadratic formula, or any other method.

Step 5: Use the Quadratic Formula

Apply the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=2\), \(b=1\), and \(c=-6\). Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4(2)(-6) = 49\).

Step 6: Calculate the Roots

Plug the values into the quadratic formula to find the roots: \[u = \frac{-1 \pm \sqrt{49}}{4}\] \[u = \frac{-1 \pm 7}{4}\] Thus, the roots are \(u = \frac{6}{4} = \frac{3}{2}\) and \(u = \frac{-8}{4} = -2\).

Step 7: Substitute Back to Solve for x

Recall \(u = e^x\). Solve \(e^x = \frac{3}{2}\) and \(e^x = -2\).\(e^x = -2\) has no valid solution as the exponential function is always positive.

Step 8: Solve for x from Valid Equation

For \(e^x = \frac{3}{2}\), take the natural logarithm to solve for \(x\). \(x = \ln\left(\frac{3}{2}\right)\).

Key Concepts

Exponential Functions

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the problem, the exponential function is expressed as \( e^{x} \), where \( e \) is Euler’s number, approximately 2.71828. Exponential functions have unique properties:

• The base \( e \) is a constant, leading to continuous growth or decay dynamics.
• They grow rapidly due to the variable exponent, which changes the base's magnitude.
• They are always positive. This is important because it helps determine the domain and range of the problem.

In this exercise, recognizing that \( 2e^{2x} + e^x = 6 \) is in terms of \( e^x \) allows us to utilize quadratic methods. The substitution method simplifies this equation, making it easier to solve.

Substitution Method

The substitution method is a technique used to transform complex equations into simpler forms. In this problem, the substitution \( u = e^x \) transforms the original equation into \( 2u^2 + u = 6 \). Here’s how substitution simplifies solving:

• By letting \( u = e^x \), we turn an exponential equation into a quadratic one, \( 2u^2 + u - 6 = 0 \), which is easier to handle.
• The main goal of substitution is to simplify the equation into a form where established methods, like the quadratic formula, can be applied.
• After solving for \( u \), we revert the substitution to find the original variable, \( x \).

This method is especially helpful in transforming complex functions into a familiar algebraic format, facilitating straightforward solutions to otherwise complicated expressions.

Quadratic Formula

The quadratic formula provides an exact way to solve quadratic equations. It is defined as \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This method is valuable because:

• It guarantees a solution when the equation is in standard quadratic form \( au^2 + bu + c = 0 \).
• The formula encompasses all potential cases for the roots (real and complex numbers).
• It directly utilizes coefficients \( a \), \( b \), and \( c \), making it applicable when factoring is difficult or impossible.

In this exercise, using the quadratic formula to solve \( 2u^2 + u - 6 = 0 \), we find \( u = \frac{-1 \pm 7}{4} \), which results in roots \( u = \frac{3}{2} \) and \( u = -2 \). By knowing these roots, we can advance in finding solutions in the original problem context.

Discriminant

The discriminant is a part of the quadratic formula, \( b^2 - 4ac \), and it reveals crucial information about a quadratic equation's roots.

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it's zero, there is exactly one real root, meaning the graph of the equation touches the x-axis at one point.
• If negative, there are no real roots, and the solutions to the equation will be complex numbers.

In this exercise, calculating the discriminant: \( 1^2 - 4(2)(-6) = 49 \) indicates two real roots. This positive result supports the applicability of the quadratic formula in obtaining \( u = \frac{3}{2} \) and \( u = -2 \). The discriminant helps verify the nature of solutions before detailed calculations, offering clarity on expected outcomes.