1. **State the problem:** Solve the equation $$\frac{e^{0.0227t}}{e^{-0.0346t}} = 2.5$$ for $t$. 2. **Use the properties of exponents:** Recall that $$\frac{e^a}{e^b} = e^{a-b}$$. 3. **Apply the property:** $$\frac{e^{0.0227t}}{e^{-0.0346t}} = e^{0.0227t - (-0.0346t)} = e^{0.0227t + 0.0346t} = e^{0.0573t}$$ 4. **Rewrite the equation:** $$e^{0.0573t} = 2.5$$ 5. **Take the natural logarithm of both sides:** $$\ln\left(e^{0.0573t}\right) = \ln(2.5)$$ 6. **Simplify using $\ln(e^x) = x$:** $$0.0573t = \ln(2.5)$$ 7. **Solve for $t$:** $$t = \frac{\ln(2.5)}{0.0573}$$ 8. **Calculate the numerical value:** $$\ln(2.5) \approx 0.9163$$ $$t \approx \frac{0.9163}{0.0573} \approx 15.99$$ **Final answer:** $$t \approx 15.99$$