1. **Problem a:** Solve $\ln(x^2 - 1) = 1$ in $\mathbb{R}$. 2. The natural logarithm function $\ln(y)$ is defined for $y > 0$. So, first ensure the argument $x^2 - 1 > 0$ which means $x^2 > 1$ or $x < -1$ or $x > 1$. 3. Use the property of logarithms: if $\ln(a) = b$, then $a = e^b$. Here, $x^2 - 1 = e^1 = e$. 4. Solve the quadratic equation: $$x^2 - 1 = e \implies x^2 = e + 1$$ 5. Taking square roots: $$x = \pm \sqrt{e + 1}$$ 6. Check domain restrictions: $x < -1$ or $x > 1$. Since $\sqrt{e+1} > 1$, both $x = \sqrt{e+1}$ and $x = -\sqrt{e+1}$ satisfy the domain. 7. **Final answer:** $$x = \pm \sqrt{e + 1}$$