To solve the equation $e^x + e^{-x} = 4$, we can follow these steps:

Step 1: Use a substitution.

Let’s define a new variable: $y = e^x$ Thus, $e^{-x} = \frac{1}{y}$. The equation becomes: $y + \frac{1}{y} = 4$

Step 2: Multiply through by $y$ to eliminate the fraction.

$y^2 + 1 = 4y$ Rearranging the equation: $y^2 - 4y + 1 = 0$

Step 3: Solve this quadratic equation.

Use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -4$, and $c = 1$. $y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)}$ $y = \frac{4 \pm \sqrt{16 - 4}}{2}$ $y = \frac{4 \pm \sqrt{12}}{2}$ $y = \frac{4 \pm 2\sqrt{3}}{2}$ $y = 2 \pm \sqrt{3}$

Thus, $y = 2 + \sqrt{3}$ or $y = 2 - \sqrt{3}$.

Step 4: Recall that $y = e^x$.

So we have: $e^x = 2 + \sqrt{3} \quad \text{or} \quad e^x = 2 - \sqrt{3}$

Step 5: Solve for $x$.

• For $e^x = 2 + \sqrt{3}$: $x = \ln(2 + \sqrt{3})$
• For $e^x = 2 - \sqrt{3}$, this is not valid because $e^x$ is always positive, and $2 - \sqrt{3}$ is less than 1, so there is no solution for this case.

Final Answer:

$x = \ln(2 + \sqrt{3})$

Would you like more details on any of the steps, or have questions about this solution?

---

Here are 5 related questions:

1. How do you solve similar equations involving $e^x + e^{-x}$?
2. What is the graph of $e^x + e^{-x}$?
3. How can you use logarithms to solve equations with exponents?
4. What are the properties of the exponential function $e^x$?
5. How do you solve a quadratic equation using the quadratic formula?

Tip: When solving exponential equations, using substitution can simplify the process and help convert them into more familiar forms like quadratic equations.