1. Solve for $x$ in the equation $$\frac{-12\sigma^{12}}{x^{13}} + \frac{6\sigma^{6}}{x^{7}} = 0.$$ 2. Start by isolating terms: $$\frac{-12\sigma^{12}}{x^{13}} = -\frac{6\sigma^{6}}{x^{7}}.$$ 3. Multiply both sides by $x^{13}$ to clear denominators: $$-12\sigma^{12} = -6\sigma^{6} x^{6}.$$ 4. Divide both sides by $-6\sigma^{6}$: $$\cancel{-12\sigma^{12}} \div \cancel{-6\sigma^{6}} = \cancel{-6\sigma^{6} x^{6}} \div \cancel{-6\sigma^{6}} \implies 2\sigma^{6} = x^{6}.$$ 5. Take the sixth root of both sides: $$x = \pm \sqrt[6]{2\sigma^{6}} = \pm \sigma \sqrt[6]{2}.$$ --- 6. Solve for $x$ in the transcendental equation $$e^{-x} = x$$ to one decimal place. 7. This equation cannot be solved algebraically, so we use graphical or numerical methods. 8. Graph $y = e^{-x}$ and $y = x$ and find their intersection. 9. By inspection or using numerical methods (e.g., Newton's method), the solution is approximately $$x \approx 0.6.$$ --- Final answers: - For the first equation: $$x = \pm \sigma \sqrt[6]{2}.$$ - For the second equation: $$x \approx 0.6.$$