Problem 85 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The exact solution is \( x = \ln(-1 + \sqrt{3}) \).

Step 1: Substitution

Let \( u = e^x \). Then \( e^{2x} = u^2 \). Substitute into the equation to get \( \frac{1}{2} u^2 + u = 1 \).

Step 2: Clear Fraction

Multiply the entire equation \( \frac{1}{2} u^2 + u = 1 \) by 2 to eliminate the fraction: \( u^2 + 2u = 2 \).

Step 3: Form a Quadratic Equation

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

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 2 \), and \( c = -2 \). Compute: \[ u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \] \[ u = \frac{-2 \pm \sqrt{4 + 8}}{2} \] \[ u = \frac{-2 \pm \sqrt{12}}{2} \] \[ u = \frac{-2 \pm 2\sqrt{3}}{2} \] \[ u = -1 \pm \sqrt{3} \].

Step 5: Back Substitute to Solve for x

Since \( u = e^x \), substitute back to get \( e^x = -1 + \sqrt{3} \) and \( e^x = -1 - \sqrt{3} \). The term \(-1 - \sqrt{3} \) is negative, which is not possible for \( e^x \). Therefore, only \( e^x = -1 + \sqrt{3} \) is valid. Solve for \( x \) by taking the natural logarithm: \[ x = \ln(-1 + \sqrt{3}) \].

Step 6: Verify the Solution

Check if the valid solution satisfies the original equation. Substitute \( x = \ln(-1 + \sqrt{3}) \) back into the original equation to ensure it holds true, confirming it as a valid solution.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). To find the solutions of such equations, we can apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula helps in not only finding real roots but also complex roots if the discriminant \( b^2 - 4ac \) is negative. Here's how it works:

• The term \( -b \) shifts the vertex of the parabola formed by the quadratic equation.
• \( \pm \sqrt{b^2 - 4ac} \) calculates the roots' spread, adjusting for the direction and distance from the vertex.
• The denominator \( 2a \) normalizes the results over the parabola's width.

In the given solution, your quadratic equation was \( u^2 + 2u - 2 = 0 \), with \( a = 1 \), \( b = 2 \), and \( c = -2 \). You use the formula to find the values of \( u \), which later aid in solving the original exponential equation.

Exponential Equations

Exponential equations involve expressions where variables appear as exponents. Often found in representing growth and decay processes, these equations can look like \( e^x \) where \( e \) is the mathematical constant approximately equal to 2.71828. Solving exponential equations cleverly involves transforming them into simpler forms or using logarithms. Exponential equations, such as \( \frac{1}{2} e^{2x} + e^x = 1 \), can be daunting initially. Here, the substitution method is particularly useful. By setting \( u = e^x \), the problem transforms into the more manageable quadratic-like form: \( \frac{1}{2}u^2 + u = 1 \). Once solved for \( u \), you then backtrace to find \( x \) through log operations, since \( \, u = e^x \). Note, not all solutions may be valid if they result in a negative value for \( e^x \), which is impossible as it must always be positive.

Substitution Method

The substitution method is an elegant algebraic technique that simplifies complex equations. By replacing a cumbersome expression with a new variable, you reduce difficulty, making it easier to identify and solve standard equations like quadratics. The idea is straightforward:

• Identify a part of the equation that can be expressed as a different variable.
• Substitute this new variable into the equation, effectively changing the problem's look.
• Once the new, simpler equation is solved, substitute back the original expression for your new variable to find the final answer.

In our original exercise, \( e^x \) was replaced by \( u \). This turned the tricky exponential equation into a quadratic equation in \( u \), which can be easily solved with the quadratic formula. Instead of wrestling with exponentials directly, substitution clarified the problem’s path forward. Remember that substitution is not just limited to exponentials and can be applied across various mathematical situations to bring clarity.