xponential Models of Growth and Decay and Average Rate of Change 0) Pika populations are increasing by \( 2.5 \% \) each year. As of 2025 , the population is 200 . How does the decrease in the pika population from years 1 to 5 compare to the decrease from years 6 to 10 ? - Write the exponential functions for pika population:

Let \( t \) represent the number of years after 2025. Because the population increases by \(2.5\%\) each year and the population in 2025 is 200, we have \[ P(t)=200\cdot (1.025)^t, \] or equivalently, for any year \( y \), \[ P(y)=200\cdot (1.025)^{y-2025}. \] Next, we compare the absolute change in population over two intervals. Although the population is increasing, we can look at how the change (difference in population) over years 1–5 compares to the change over years 6–10. 1. For years 1 to 5 (i.e. \( t=1 \) to \( t=5 \)): \[ \Delta P_{1\to5}=P(5)-P(1)=200\cdot (1.025)^5-200\cdot (1.025)^1. \] 2. For years 6 to 10 (i.e. \( t=6 \) to \( t=10 \)): \[ \Delta P_{6\to10}=P(10)-P(6)=200\cdot (1.025)^{10}-200\cdot (1.025)^6. \] Because exponential functions grow at an increasing rate, we expect \[ \Delta P_{6\to10} > \Delta P_{1\to5}. \] To see this more clearly, factor as follows: - For years 1 to 5: \[ \Delta P_{1\to5}=200\cdot\Bigl[(1.025)^5-(1.025)^1\Bigr]. \] - For years 6 to 10: \[ \Delta P_{6\to10}=200\cdot\Bigl[(1.025)^{10}-(1.025)^6\Bigr]. \] Notice that \[ (1.025)^{10}=(1.025)^5\cdot(1.025)^5 \quad \text{and} \quad (1.025)^6=(1.025)^1\cdot(1.025)^5. \] So we can write \[ \Delta P_{6\to10}=200\cdot (1.025)^5\Bigl[(1.025)^5-(1.025)^1\Bigr]. \] This shows that \[ \Delta P_{6\to10}=(1.025)^5\cdot\Delta P_{1\to5}. \] Since \((1.025)^5>1\), the change (increase) in the later interval (years 6 to 10) is larger than the change in the earlier interval (years 1 to 5).