You are given the equation:

\[
\frac{e^x + e^{-x}}{e^x - e^{-x}} = 2
\]

Let’s solve this step by step.

### Step 1: Define substitutions
Let us define \( y = e^x \). Then \( e^{-x} = \frac{1}{y} \). Substituting these into the equation:

\[
\frac{y + \frac{1}{y}}{y - \frac{1}{y}} = 2
\]

### Step 2: Simplify the numerator and denominator
First, simplify the numerator and denominator separately:

Numerator:
\[
y + \frac{1}{y} = \frac{y^2 + 1}{y}
\]

Denominator:
\[
y - \frac{1}{y} = \frac{y^2 - 1}{y}
\]

Now the equation becomes:
\[
\frac{\frac{y^2 + 1}{y}}{\frac{y^2 - 1}{y}} = 2
\]

### Step 3: Simplify the equation
Since both the numerator and the denominator have \( \frac{1}{y} \), they cancel out:
\[
\frac{y^2 + 1}{y^2 - 1} = 2
\]

### Step 4: Cross multiply
Cross multiply to get rid of the fraction:
\[
y^2 + 1 = 2(y^2 - 1)
\]

### Step 5: Expand and simplify
Expand both sides:
\[
y^2 + 1 = 2y^2 - 2
\]

Now, move all terms to one side:
\[
1 + 2 = 2y^2 - y^2
\]
\[
3 = y^2
\]

### Step 6: Solve for \( y \)
Taking the square root of both sides:
\[
y = \pm \sqrt{3}
\]

### Step 7: Solve for \( x \)
Recall that \( y = e^x \), so:
\[
e^x = \pm \sqrt{3}
\]

For \( e^x = \sqrt{3} \), take the natural logarithm on both sides:
\[
x = \ln(\sqrt{3}) = \frac{1}{2} \ln(3)
\]

However, \( e^x \) cannot be negative, so we discard the negative solution \( e^x = -\sqrt{3} \).

### Final solution:
\[
x = \frac{1}{2} \ln(3)
\]

Would you like further details or have any questions?

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Here are 5 related questions to expand this:

1. How would the solution change if the equation were \( \frac{e^x - e^{-x}}{e^x + e^{-x}} = 2 \)?
2. How do we apply logarithmic differentiation in this type of equation?
3. What happens if we substitute \( e^x = 2 \) in similar equations?
4. How would the result change if we used base 10 logarithms instead of natural logarithms?
5. How can we solve equations involving both exponential and logarithmic terms?

**Tip:** When solving exponential equations, logarithms are powerful tools to isolate variables in exponents.

Learn how to solve the exponential equation e^x + e^-x / e^x - e^-x = 2 step by step. The solution involves algebraic manipulation, substitution, and the application of logarithmic functions. Ideal for students in Grades 10-12 studying advanced algebra and exponential functions.

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