1. **State the problem:** Solve the equation $$e^{-\ln x} = 20$$ for $x$. 2. **Recall the properties:** The natural logarithm and exponential functions are inverses, so $$e^{\ln a} = a$$ for $a > 0$. 3. **Rewrite the expression:** Using the property of exponents, $$e^{-\ln x} = \frac{1}{e^{\ln x}}$$. 4. **Simplify:** Since $$e^{\ln x} = x$$, we have $$e^{-\ln x} = \frac{1}{x}$$. 5. **Set up the equation:** $$\frac{1}{x} = 20$$. 6. **Solve for $x$:** Multiply both sides by $x$ and divide by 20: $$x = \frac{1}{20}$$. 7. **Check domain:** Since $x$ must be positive (domain of $\ln x$), $x = \frac{1}{20}$ is valid. **Final answer:** $$x = \frac{1}{20}$$