1. **State the problem:** Solve the equation $$\frac{e^x - e^{-x}}{2} = 1$$ for $x$. 2. **Rewrite the equation:** Multiply both sides by 2 to clear the denominator: $$e^x - e^{-x} = 2$$ 3. **Substitute:** Let $u = e^x$. Then $e^{-x} = \frac{1}{u}$. The equation becomes: $$u - \frac{1}{u} = 2$$ 4. **Clear the fraction:** Multiply both sides by $u$: $$u^2 - 1 = 2u$$ 5. **Rearrange into quadratic form:** $$u^2 - 2u - 1 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$u = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$ 7. **Select valid solution:** Since $u = e^x > 0$, discard $1 - \sqrt{2}$ (negative). So: $$e^x = 1 + \sqrt{2}$$ 8. **Solve for $x$:** Take natural logarithm on both sides: $$x = \ln(1 + \sqrt{2})$$ **Final answer:** $$x = \ln(1 + \sqrt{2})$$