1. **State the problem:** We have investment values $V(n)$ at different years $n$ and want to find an exponential regression model of the form $$V(n) = a \cdot b^n$$ where $a$ is the initial value and $b$ is the growth factor per year. 2. **Use the data points:** Given $n = 0,3,7,12,14,19$ and corresponding $V(n) = 2400, 2722.73, 3377.04, 4439.36, 4751.84, 5761.45$. 3. **Find $a$ and $b$ using exponential regression:** Since $V(0) = a \cdot b^0 = a$, initial value $a = 2400$. 4. **Calculate $b$ using two points, for example $n=0$ and $n=19$:** $$b = \left(\frac{V(19)}{V(0)}\right)^{\frac{1}{19}} = \left(\frac{5761.45}{2400}\right)^{\frac{1}{19}} = (2.4006)^{0.05263} \approx 1.047$$ 5. **Exponential regression equation:** $$V(n) = 2400 \times 1.047^n$$ 6. **Percent increase per year:** Percent increase $= (b - 1) \times 100 = (1.047 - 1) \times 100 = 4.7\%$ per year. 7. **Find $V(20)$:** $$V(20) = 2400 \times 1.047^{20} = 2400 \times 2.512 = 6028.80$$ 8. **Interpretation:** "The value of the investment after 20 years is approximately 6028.80 dollars." 9. **Find time to reach 10,000:** Solve for $n$ in $$10000 = 2400 \times 1.047^n$$ Divide both sides by 2400: $$\frac{10000}{2400} = 1.047^n \Rightarrow 4.1667 = 1.047^n$$ Take natural log: $$\ln(4.1667) = n \ln(1.047)$$ $$n = \frac{\ln(4.1667)}{\ln(1.047)} = \frac{1.427}{0.046} \approx 31.02$$ 10. **Interpretation:** "In approximately 31.02 years, the value of the investment will reach 10000 dollars." 11. **Find time to double the investment:** Solve for $n$ in $$2 \times 2400 = 2400 \times 1.047^n$$ $$2 = 1.047^n$$ Take natural log: $$\ln(2) = n \ln(1.047)$$ $$n = \frac{\ln(2)}{\ln(1.047)} = \frac{0.693}{0.046} \approx 15.07$$ 12. **Interpretation:** "In approximately 15.07 years, the value of the investment will double."