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?

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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.

Discover how to solve the exponential equation e^x + e^−x = 4 step-by-step by using substitution and the quadratic formula. This problem covers important concepts such as exponential and quadratic equations, as well as the application of logarithms. Ideal for students in Grades 10-12.

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