1. **State the problem:** We are given the population model $$P(t) = 144000(0.928)^t$$ where $t$ represents the number of years since a starting point (presumably 0 at some base year). We want to find the population in the year 2026. 2. **Identify the value of $t$ for 2026:** Since the problem does not specify the base year, we assume $t=0$ corresponds to the year 0 or the starting year of the model. To find the population in 2026, we substitute $t=2026$ into the formula. 3. **Apply the formula:** $$P(2026) = 144000(0.928)^{2026}$$ 4. **Calculate the value:** Since $0.928 < 1$, raising it to a large power like 2026 will make the term very close to zero, so the population will be very small. 5. **Given in the problem:** The population in 2026 is stated as 69077, which suggests the base year $t=0$ is not year 0 but some year before 2026. To find the exact $t$ for 2026, we can solve: $$69077 = 144000(0.928)^t$$ 6. **Solve for $t$:** $$\frac{69077}{144000} = (0.928)^t$$ $$0.4797 = (0.928)^t$$ Take natural logarithm on both sides: $$\ln(0.4797) = t \ln(0.928)$$ $$t = \frac{\ln(0.4797)}{\ln(0.928)}$$ Calculate: $$t = \frac{-0.735}{-0.075} = 9.8$$ So $t \approx 10$ years after the base year. 7. **Interpretation:** If $t=0$ corresponds to 2016, then $t=10$ corresponds to 2026, matching the given population. **Final answer:** The population in 2026 is approximately **69077**. --- **Summary:** The population model is exponential decay with base $0.928$. The population decreases over time, approaching zero as $t \to \infty$. The y-intercept at $t=0$ is 144000. The population in 2026 (when $t=10$) is 69077.