1. **Problem statement:** Solve for $x$ in the equation $$e^{x-1} - 18e^{-x} - 3 = 0.$$ 2. **Rewrite the equation:** Recall that $e^{-x} = \frac{1}{e^x}$. So the equation becomes $$e^{x-1} - 18 \frac{1}{e^x} - 3 = 0.$$ 3. **Express terms with $e^x$:** Note that $e^{x-1} = e^x \cdot e^{-1} = \frac{e^x}{e}$. Substitute this to get $$\frac{e^x}{e} - \frac{18}{e^x} - 3 = 0.$$ 4. **Multiply through by $e^x$ to clear denominators:** $$e^x \cdot \frac{e^x}{e} - 18 - 3 e^x = 0 \implies \frac{(e^x)^2}{e} - 3 e^x - 18 = 0.$$ 5. **Let $t = e^x$ (note $t > 0$):** The equation becomes $$\frac{t^2}{e} - 3t - 18 = 0.$$ Multiply both sides by $e$ to clear denominator: $$t^2 - 3 e t - 18 e = 0.$$ 6. **Solve quadratic in $t$:** Using quadratic formula $$t = \frac{3 e \pm \sqrt{(3 e)^2 + 4 \cdot 18 e}}{2} = \frac{3 e \pm \sqrt{9 e^2 + 72 e}}{2}.$$ 7. **Simplify the discriminant:** $$\sqrt{9 e^2 + 72 e} = \sqrt{9 e (e + 8)} = 3 \sqrt{e (e + 8)}.$$ 8. **So, $$t = \frac{3 e \pm 3 \sqrt{e (e + 8)}}{2} = \frac{3 e}{2} \pm \frac{3}{2} \sqrt{e (e + 8)}.$$** 9. **Since $t = e^x > 0$, both roots must be checked for positivity:** - Root 1: $$t_1 = \frac{3 e}{2} + \frac{3}{2} \sqrt{e (e + 8)} > 0.$$ - Root 2: $$t_2 = \frac{3 e}{2} - \frac{3}{2} \sqrt{e (e + 8)}.$$ Since $\sqrt{e (e + 8)} > e$, $t_2$ is negative, discard. 10. **Find $x$ from $t_1$: $$x = \ln t_1 = \ln \left( \frac{3 e}{2} + \frac{3}{2} \sqrt{e (e + 8)} \right).$$** **Final answer:** $$\boxed{x = \ln \left( \frac{3 e}{2} + \frac{3}{2} \sqrt{e (e + 8)} \right)}.$$