1. **State the problem:** We are given the exponential population model $$A=886.3e^{0.019t}$$ where $A$ is the population in millions and $t$ is the number of years after 2003. We need to find the value of $t$ when the population $A$ reaches 1092 million. 2. **Write the equation to solve:** Set $A=1092$ and solve for $t$: $$1092 = 886.3e^{0.019t}$$ 3. **Isolate the exponential term:** Divide both sides by 886.3: $$\frac{1092}{886.3} = e^{0.019t}$$ 4. **Simplify the fraction:** $$1.232 = e^{0.019t}$$ 5. **Take the natural logarithm of both sides:** $$\ln(1.232) = \ln\left(e^{0.019t}\right)$$ 6. **Use the logarithm power rule:** $$\ln(1.232) = 0.019t$$ 7. **Solve for $t$:** $$t = \frac{\ln(1.232)}{0.019}$$ 8. **Calculate the value:** $$t \approx \frac{0.2087}{0.019} \approx 10.98$$ 9. **Interpret the result:** Since $t$ is years after 2003, the population will reach 1092 million approximately 11 years after 2003, which is in the year 2014. **Final answer:** The population will be 1092 million in approximately **11 years** after 2003, i.e., in 2014.