Problem 86 - A Graphical Approach to College Algebra

Question

$$\text { The given equations are quadratic in form. Solve each and give exact solutions.}$$ $$\frac{1}{4} e^{2 x}+2 e^{x}=3$$

Step-by-Step Solution

Answer

The exact solution is \( x = \ln(-4 + 2\sqrt{7}) \).

Step 1: Change of Variables

Recognize that this equation is quadratic in form. To make it simpler, use a substitution. Let \( u = e^{x} \), then \( u^2 = e^{2x} \). The equation becomes: \[ \frac{1}{4} u^2 + 2u = 3 \].

Step 2: Eliminate Fractions

To eliminate the fraction, multiply the entire equation by 4 to simplify. You get: \[ u^2 + 8u = 12 \].

Step 3: Rearrange to Standard Quadratic Form

Rearrange your equation to get a standard quadratic equation. Subtract 12 from both sides: \[ u^2 + 8u - 12 = 0 \].

Step 4: Solve the Quadratic Equation

Solve this quadratic equation. Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 8, c = -12 \). Substitute and solve: \[ u = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \].

Step 5: Simplify the Solution

Calculate the discriminant and solve for \( u \): \[ u = \frac{-8 \pm \sqrt{64 + 48}}{2} = \frac{-8 \pm \sqrt{112}}{2} = \frac{-8 \pm 4\sqrt{7}}{2} \].This simplifies to \[ u = -4 \pm 2\sqrt{7} \].

Step 6: Reverse the Substitution

Since \( u = e^{x} \), find \( x \) for the values of \( u \). We have two cases:1. \( e^{x} = -4 + 2\sqrt{7} \)2. \( e^{x} = -4 - 2\sqrt{7} \).The second case is impossible since \( e^{x} \) must be positive.

Step 7: Solve for x

For \( e^{x} = -4 + 2\sqrt{7} \), we take the natural logarithm:\( x = \ln(-4 + 2\sqrt{7}) \) being the exact solution assuming \( -4 + 2\sqrt{7} > 0 \). Calculate \( 2\sqrt{7} \approx 5.29 \), so \( -4 + 5.29 > 0 \) is valid.

Key Concepts

Exponential Equations

Understanding exponential equations is crucial, as they frequently appear in various math and science problems. An exponential equation is one where variables appear in the exponent. In our exercise, \( e^{2x} \) and \( e^{x} \) represent exponential expressions.
These types of equations often require special techniques to solve, such as taking logarithms or using substitutions.

• Exponential equations can sometimes be transformed into simpler forms using algebraic techniques, such as substituting variables, which helps in solving them as if they were polynomial equations.
• It is essential to remember that exponential functions are always positive, hence any solutions must reflect this fundamental property.

A common method to solve exponential equations is using the change of variables, where we replace complex exponential terms with a simpler variable, making the equation more manageable.

Change of Variables

The technique of change of variables simplifies complex equations by substituting a difficult-to-handle expression with a single variable.
In the original exercise, because the equation \( \frac{1}{4} e^{2x} + 2 e^{x} = 3 \) contains exponential terms, we use a change of variables to transform it into a quadratic form.

• Here, let \( u = e^{x} \), so \( u^2 = e^{2x} \).
• This transforms the equation into a new form: \( \frac{1}{4} u^2 + 2u = 3 \), which is easier to solve.

After substituting, it is often necessary to further rearrange the equation to conform to standard forms that we know how to solve, such as the quadratic form. This technique not only simplifies calculations but also helps identify hidden structures within equations, often pivotal in finding solutions.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. The standard quadratic equation is expressed as \( ax^2 + bx + c = 0 \).
The quadratic formula provides the solutions for \( x \) as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

• The discriminant, \( b^2 - 4ac \), plays a crucial role in determining the nature of the solutions.
• If the discriminant is positive, there are two distinct real solutions. A zero discriminant indicates a repeated real solution, while a negative discriminant means there are no real solutions, only complex ones.

In the original problem, after substituting and transforming into a quadratic form: \( u^2 + 8u - 12 = 0 \), we apply the quadratic formula to find \( u \). Using \( a = 1 \), \( b = 8 \), and \( c = -12 \), we calculate the solutions for \( u \) and then back-substitute to solve for \( x \).Utilizing the quadratic formula not only finds exact solutions efficiently but also ensures that all possible solutions are explored systematically. Understanding this formula is essential for tackling complex algebraic equations beyond simple quadratics.