1. **State the problem:** We have population data for years 2004, 2008, 2012, 2016, and 2020, with populations in millions: 22.4, 24.3, 26.0, 27.9, and 29.2 respectively. 2. The model is an exponential function of the form $$f(x) = a \cdot b^x$$ where $x=4$ corresponds to 2004, $x=8$ to 2008, etc. 3. We want to find the year when the population first reaches 35 million, i.e., solve for $x$ in $$f(x) = 35$$. 4. **Find parameters $a$ and $b$ using two points:** Using $x=4$, $f(4)=22.4$ and $x=8$, $f(8)=24.3$: $$22.4 = a \cdot b^4$$ $$24.3 = a \cdot b^8$$ 5. Divide the second equation by the first to eliminate $a$: $$\frac{24.3}{22.4} = \frac{a \cdot b^8}{a \cdot b^4} = b^{8-4} = b^4$$ $$b^4 = \frac{24.3}{22.4} \approx 1.084\Rightarrow b = \sqrt[4]{1.084}$$ 6. Calculate $b$: $$b = 1.084^{\frac{1}{4}} \approx 1.0205$$ 7. Substitute $b$ back to find $a$ using $f(4)=22.4$: $$22.4 = a \cdot (1.0205)^4$$ $$22.4 = a \cdot 1.084$$ $$a = \frac{22.4}{1.084} \approx 20.67$$ 8. The model is: $$f(x) = 20.67 \cdot (1.0205)^x$$ 9. Solve for $x$ when $f(x) = 35$: $$35 = 20.67 \cdot (1.0205)^x$$ $$\frac{35}{20.67} = (1.0205)^x$$ $$1.693 = (1.0205)^x$$ 10. Take natural logarithm on both sides: $$\ln(1.693) = x \cdot \ln(1.0205)$$ $$x = \frac{\ln(1.693)}{\ln(1.0205)}$$ 11. Calculate $x$: $$x = \frac{0.527}{0.0203} \approx 25.96$$ 12. Recall $x=4$ corresponds to 2004, so the year is: $$2000 + x = 2000 + 25.96 = 2025.96 \approx 2026$$ 13. Since population reaches 35 million just after 2025, the first full year it reaches 35 million is 2026. **Answer:** The population first reaches 35 million in the year 2026, so the closest choice is **2025** (option A).