Problem 31 - Precalculus Mathematics for Calculus

Question

Solve the equation. $$e^{4 x}+4 e^{2 x}-21=0$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{2} \ln(3)\).

Step 1: Recognize the Substitution

Notice that the equation is in the form of a quadratic in terms of \(e^{2x}\). Let \(u = e^{2x}\). This transforms the equation into: \[u^2 + 4u - 21 = 0.\]

Step 2: Solve the Quadratic Equation

The equation \(u^2 + 4u - 21 = 0\) is a standard quadratic equation. We will use the quadratic formula, \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = -21\).

Step 3: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \[u = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-21)}}{2 \times 1}.\]Simplify the expression under the square root: \[b^2 - 4ac = 16 + 84 = 100.\]

Step 4: Find the Roots for u

Continue simplifying: \[u = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}.\]This gives us two possible values for \(u\):1. \(u = \frac{6}{2} = 3\)2. \(u = \frac{-14}{2} = -7\).

Step 5: Re-substitute for \(e^{2x}\) and Solve for x

Since \(e^{2x}\) must be a positive number, only \(u = 3\) is valid, because \(e^{2x} = -7\) has no real solutions.So, \(e^{2x} = 3\). To solve for \(x\), take the natural logarithm of both sides:\[2x = \ln(3)\]\[x = \frac{1}{2} \ln(3).\]

Key Concepts

Quadratic Substitution

When you first see an equation like \(e^{4x} + 4e^{2x} - 21 = 0\), it might look pretty intimidating because of the exponentials. But there's a handy trick called quadratic substitution that can make solving this much easier.

Here's how it works:

• Notice that the equation can be rewritten to get a form like \((e^{2x})^2\), which suggests a quadratic form.
• To simplify, set \(u = e^{2x}\). This transforms the equation into a more familiar form: \(u^2 + 4u - 21 = 0\).
• Now, you have a simpler quadratic equation in terms of \(u\), instead of dealing directly with the exponentials.
  

This substitution helps you bypass the complexity of exponential equations by converting them into quadratic ones, allowing you to use straightforward algebraic techniques to find solutions.

Natural Logarithm

After you've solved the quadratic equation and determined that \(u = 3\), you need to return to the context of the original problem by substituting back for \(e^{2x}\). This gives us \(e^{2x} = 3\).

The next step involves natural logarithms. The natural logarithm, denoted as \(\ln\), is the inverse of the exponential function with base \(e\). It's a key tool when you want to "undo" an exponential function.

• To solve \(e^{2x} = 3\) for \(x\), take the natural logarithm of both sides: \(2x = \ln(3)\).
• This step works because the logarithm and the exponential are inverse operations—like adding and subtracting, or multiplying and dividing—so they essentially "cancel each other out."
• This tells you that \(2x\) equals the natural logarithm of 3.
  

Finally, remember to solve for \(x\) by dividing both sides by 2, resulting in \(x = \frac{1}{2}\ln(3)\). This step brings you back from the substitution to your original variable, \(x\).

Solving Equations

Solving equations is a fundamental skill in algebra, and it involves finding the values of variables that make an equation true. In this exercise, we applied several key techniques to solve the given equation.

• We started with a seemingly complex exponential equation and identified that it could be simplified by recognizing its quadratic form.
• Utilizing quadratic substitution, we converted the exponential equation into a quadratic equation, which is easier to manage.
• After solving the quadratic equation using the quadratic formula, the work wasn't over. We had to return the substituted variable \(u\) back to its original form \(e^{2x}\).
• Given \(e^{2x} = 3\), we used the natural logarithm to solve for \(x\). This final step required understanding how logarithms and exponentials interact as inverse functions.

These strategies illustrate that often the path to solving more complex equations involves simplifying and transforming them using known algebraic techniques. Each equation may involve a unique blend of methods to arrive at a solution, highlighting the creativity and logic at the heart of algebra.