1. The problem states: If $\cosh(x) = 2$, find the value of $x$. 2. Recall the definition of hyperbolic cosine: $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$ 3. Set up the equation: $$\frac{e^x + e^{-x}}{2} = 2$$ 4. Multiply both sides by 2: $$e^x + e^{-x} = 4$$ 5. Let $y = e^x$, then $e^{-x} = \frac{1}{y}$, so: $$y + \frac{1}{y} = 4$$ 6. Multiply both sides by $y$ to clear the fraction: $$y^2 + 1 = 4y$$ 7. Rearrange to form a quadratic equation: $$y^2 - 4y + 1 = 0$$ 8. Solve the quadratic using the quadratic formula: $$y = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$ 9. Recall $y = e^x$, so: $$e^x = 2 \pm \sqrt{3}$$ 10. Take the natural logarithm of both sides: $$x = \ln(2 \pm \sqrt{3})$$ 11. Therefore, the solutions are: $$x = \ln(2 + \sqrt{3}) \quad \text{or} \quad x = \ln(2 - \sqrt{3})$$ 12. Among the options, this corresponds to option b: $\ln(2 \pm \sqrt{3})$. Final answer: $x = \ln(2 \pm \sqrt{3})$