Problem 31 - Algebra and Trigonometry

Question

Solve the equation. $$ e^{2 x}-3 e^{x}+2=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = \ln(2) \).

Step 1: Substitution

Recognize that the equation is quadratic in form. Let \( u = e^x \). Then \( e^{2x} = (e^x)^2 = u^2 \), so the equation becomes \( u^2 - 3u + 2 = 0 \).

Step 2: Factoring the Quadratic

Factor the quadratic equation \( u^2 - 3u + 2 = 0 \). The equation can be factored as \((u - 1)(u - 2) = 0\).

Step 3: Solve for u

Solve the equation \((u - 1)(u - 2) = 0\) by setting each factor equal to zero. Thus, \(u - 1 = 0\) or \(u - 2 = 0\), giving us \(u = 1\) or \(u = 2\).

Step 4: Back-Substitution

Recall that \( u = e^x \). Substitute back to find \( e^x = 1 \) and \( e^x = 2 \).

Step 5: Solve for x

Solve each equation from Step 4. For \( e^x = 1 \), take the natural log: \( x = \ln(1) = 0 \). For \( e^x = 2 \), take the natural log: \( x = \ln(2) \).

Step 6: Conclusion

The solutions to the original equation \( e^{2x} - 3e^x + 2 = 0 \) are \( x = 0 \) and \( x = \ln(2) \).

Key Concepts

Quadratic Form

Many exponential equations can take the form of a quadratic equation, which is familiar and easier to solve. For example, in the equation given, \(e^{2x} - 3e^x + 2 = 0\), it may not initially seem like a quadratic equation, but it can be transformed into one. By making a substitution, such as letting \(u = e^x\), we convert our exponential terms into simple powers of \(u\). Now the equation looks like \(u^2 - 3u + 2 = 0\), which is a classic quadratic equation.
This transformation is often known as 'reducing an equation to quadratic form.' It leverages the structure of the original problem to create an equation that is solvable with familiar techniques of factoring, applying the quadratic formula, or completing the square.

Natural Logarithm

The natural logarithm, represented by \(\ln\), is a valuable tool when dealing with exponential equations. After simplifying an exponential equation to a form such as \(e^x = a\), the natural logarithm comes in handy.

• The natural logarithm, \(\ln(a)\), is equivalent to asking, "To what power must \(e\) be raised to yield \(a\)?"
• For example, if \(e^x = 1\), then taking the natural logarithm of both sides gives \(x = \ln(1) = 0\) since \(e^0 = 1\).
• Similarly, if \(e^x = 2\), it follows that \(x = \ln(2)\).

This process is crucial because it allows us to find the value of \(x\) in terms of known constants, providing exact solutions for even initially complex equations.

Back-Substitution

Back-substitution is a method used after simplifying an equation through substitution. Once we've solved the simplified equation, we need to return to our original variable to find the solution in terms of it.
In the example \(e^{2x} - 3e^x + 2 = 0\), we initially let \(u = e^x\) to transform the equation into a solvable form. After solving the quadratic \(u^2 - 3u + 2 = 0\) and finding \(u = 1\) and \(u = 2\), we perform back-substitution by replacing \(u\) with \(e^x\) again. Thus, we have the equations \(e^x = 1\) and \(e^x = 2\).
This technique ensures that we interpret solutions in the context of the original problem, making our final answer relevant and accurate.