Exponential Growth And Decay Practice Problems Hot Sale | www ... - Free Printable

Problem 1: Population of Ontario

The population of Ontario is modeled by the function:
\[ P = 10.1(1.0125)^n \]
where \( n \) is the number of years since 1991.

#### a) Estimate the population of Ontario in:
- i) 1991
- For 1991, \( n = 0 \):
\[
P = 10.1(1.0125)^0 = 10.1 \times 1 = 10.1 \text{ million}
\]

- ii) 2000
- For 2000, \( n = 2000 - 1991 = 9 \):
\[
P = 10.1(1.0125)^9
\]
Using a calculator:
\[
(1.0125)^9 \approx 1.12617
\]
\[
P \approx 10.1 \times 1.12617 \approx 11.3743 \text{ million}
\]

- iii) 2005
- For 2005, \( n = 2005 - 1991 = 14 \):
\[
P = 10.1(1.0125)^{14}
\]
Using a calculator:
\[
(1.0125)^{14} \approx 1.18913
\]
\[
P \approx 10.1 \times 1.18913 \approx 12.0002 \text{ million}
\]

#### b) At what rate is the population growing?
The growth rate is given by the base of the exponential function, which is \( 1.0125 \). This means the population grows by:
\[
1.0125 - 1 = 0.0125 \quad \text{or} \quad 1.25\% \text{ per year}
\]

Final Answer for Problem 1:

\[
\boxed{10.1, 11.3743, 12.0002, 1.25\%}
\]

---

Problem 2: Car Depreciation

The value of a car \( n \) years after purchase is given by:
\[ V = 20000(0.84)^n \]

#### a) What is the value of the car when it is 5 years old?
- For \( n = 5 \):
\[
V = 20000(0.84)^5
\]
Using a calculator:
\[
(0.84)^5 \approx 0.4182
\]
\[
V \approx 20000 \times 0.4182 \approx 8364 \text{ dollars}
\]

#### b) What is the value of the car after the first year?
- For \( n = 1 \):
\[
V = 20000(0.84)^1 = 20000 \times 0.84 = 16800 \text{ dollars}
\]

#### c) By what percent does the value of the car depreciate each year?
The depreciation rate is given by the base of the exponential function, which is \( 0.84 \). This means the car depreciates by:
\[
1 - 0.84 = 0.16 \quad \text{or} \quad 16\% \text{ per year}
\]

Final Answer for Problem 2:

\[
\boxed{8364, 16800, 16\%}
\]

---

Problem 3: Stamp Value

The value of a rare stamp is given by:
\[ V = 65(1.07)^n \]

#### a) What will the stamp be worth in the year 2050?
- For 2050, \( n = 2050 - 1995 = 55 \):
\[
V = 65(1.07)^{55}
\]
Using a calculator:
\[
(1.07)^{55} \approx 30.425
\]
\[
V \approx 65 \times 30.425 \approx 1977.625 \text{ dollars}
\]

#### b) What value was the stamp appraised at in 1995?
- For 1995, \( n = 0 \):
\[
V = 65(1.07)^0 = 65 \times 1 = 65 \text{ dollars}
\]

#### c) At what rate is the value increasing?
The growth rate is given by the base of the exponential function, which is \( 1.07 \). This means the value increases by:
\[
1.07 - 1 = 0.07 \quad \text{or} \quad 7\% \text{ per year}
\]

Final Answer for Problem 3:

\[
\boxed{1977.625, 65, 7\%}
\]

---

Problem 4: Blue Jeans Fading

The percent of original color remaining after \( x \) washings is given by:
\[ C = 100(0.98)^x \]

#### a) What percent of color will remain after 20 washings?
- For \( x = 20 \):
\[
C = 100(0.98)^{20}
\]
Using a calculator:
\[
(0.98)^{20} \approx 0.6676
\]
\[
C \approx 100 \times 0.6676 \approx 66.76\%
\]

#### b) What percent of color will remain after 50 washings?
- For \( x = 50 \):
\[
C = 100(0.98)^{50}
\]
Using a calculator:
\[
(0.98)^{50} \approx 0.3642
\]
\[
C \approx 100 \times 0.3642 \approx 36.42\%
\]

#### c) What percentage of color do the jeans lose after each wash?
The loss rate is given by the base of the exponential function, which is \( 0.98 \). This means the jeans lose:
\[
1 - 0.98 = 0.02 \quad \text{or} \quad 2\% \text{ per wash}
\]

Final Answer for Problem 4:

\[
\boxed{66.76\%, 36.42\%, 2\%}
\]

---

Problem 5: Bacteria Growth

The population of bacteria grows by 80% every 2 days. The initial population is 212 bacteria.

#### a) After 4 days
- Since the population grows by 80% every 2 days, after 4 days (which is 2 periods of 2 days):
\[
P = 212 \times (1 + 0.80)^2 = 212 \times (1.80)^2
\]
Using a calculator:
\[
(1.80)^2 = 3.24
\]
\[
P = 212 \times 3.24 = 686.88 \approx 687 \text{ bacteria}
\]

#### b) After 8 days
- After 8 days (which is 4 periods of 2 days):
\[
P = 212 \times (1.80)^4
\]
Using a calculator:
\[
(1.80)^4 \approx 10.4976
\]
\[
P = 212 \times 10.4976 \approx 2225.47 \approx 2225 \text{ bacteria}
\]

#### c) After 11 days
- After 11 days, there are 5 full periods of 2 days and an additional 1 day. First, calculate the population after 10 days:
\[
P_{10} = 212 \times (1.80)^5
\]
Using a calculator:
\[
(1.80)^5 \approx 18.89568
\]
\[
P_{10} = 212 \times 18.89568 \approx 4000.88 \approx 4001 \text{ bacteria}
\]
Now, calculate the growth for the additional 1 day. Since the population grows by 80% every 2 days, the daily growth rate is:
\[
\sqrt{1.80} \approx 1.3416
\]
\[
P_{11} = 4001 \times 1.3416 \approx 5366.84 \approx 5367 \text{ bacteria}
\]

Final Answer for Problem 5:

\[
\boxed{687, 2225, 5367}
\]

---

Problem 6: Termite Population

The termite population decreases by 40% every day. The initial population is 800,000 termites.

#### a) After 3 days
- The population decreases by 40% each day, so it retains 60% of its previous value each day. After 3 days:
\[
P = 800000 \times (0.60)^3
\]
Using a calculator:
\[
(0.60)^3 = 0.216
\]
\[
P = 800000 \times 0.216 = 172800 \text{ termites}
\]

#### b) After 36 hours
- 36 hours is equivalent to 1.5 days. After 1.5 days:
\[
P = 800000 \times (0.60)^{1.5}
\]
Using a calculator:
\[
(0.60)^{1.5} \approx 0.520
\]
\[
P = 800000 \times 0.520 = 416000 \text{ termites}
\]

#### c) After 2 weeks
- 2 weeks is equivalent to 14 days. After 14 days:
\[
P = 800000 \times (0.60)^{14}
\]
Using a calculator:
\[
(0.60)^{14} \approx 0.0000478
\]
\[
P = 800000 \times 0.0000478 \approx 38.24 \approx 38 \text{ termites}
\]

Final Answer for Problem 6:

\[
\boxed{172800, 416000, 38}
\]

---

Problem 7: Mealworm Population

The mealworm population doubles every month. The initial population is 10 mealworms.

#### a) After 3 months
- After 3 months:
\[
P = 10 \times 2^3 = 10 \times 8 = 80 \text{ mealworms}
\]

#### b) After 1 year
- 1 year is equivalent to 12 months. After 12 months:
\[
P = 10 \times 2^{12}
\]
Using a calculator:
\[
2^{12} = 4096
\]
\[
P = 10 \times 4096 = 40960 \text{ mealworms}
\]

#### c) After 45 days
- 45 days is approximately 1.5 months. After 1.5 months:
\[
P = 10 \times 2^{1.5}
\]
Using a calculator:
\[
2^{1.5} \approx 2.828
\]
\[
P = 10 \times 2.828 \approx 28.28 \approx 28 \text{ mealworms}
\]

Final Answer for Problem 7:

\[
\boxed{80, 40960, 28}
\]

---

Problem 8: Bacteria in Ear

The bacteria population decreases by 7% every 6 hours. The initial population is 50,000,000 bacteria.

#### a) After 1 day
- 1 day is equivalent to 4 periods of 6 hours. After 1 day:
\[
P = 50000000 \times (0.93)^4
\]
Using a calculator:
\[
(0.93)^4 \approx 0.7481
\]
\[
P = 50000000 \times 0.7481 \approx 37405000 \text{ bacteria}
\]

#### b) After 5 days
- 5 days is equivalent to 20 periods of 6 hours. After 5 days:
\[
P = 50000000 \times (0.93)^{20}
\]
Using a calculator:
\[
(0.93)^{20} \approx 0.256
\]
\[
P = 50000000 \times 0.256 \approx 12800000 \text{ bacteria}
\]

#### c) After 1 week
- 1 week is equivalent to 28 periods of 6 hours. After 1 week:
\[
P = 50000000 \times (0.93)^{28}
\]
Using a calculator:
\[
(0.93)^{28} \approx 0.138
\]
\[
P = 50000000 \times 0.138 \approx 6900000 \text{ bacteria}
\]

#### d) After 3 hours
- 3 hours is half of 6 hours. After 3 hours:
\[
P = 50000000 \times (0.93)^{0.5}
\]
Using a calculator:
\[
(0.93)^{0.5} \approx 0.9644
\]
\[
P = 50000000 \times 0.9644 \approx 48220000 \text{ bacteria}
\]

Final Answer for Problem 8:

\[
\boxed{37405000, 12800000, 6900000, 48220000}
\]

---

Problem 9: Van Gogh Painting Value

The value of a Van Gogh painting increases by 7.3% each year. The initial value in 1952 was $94,000.

#### After 2020
- The number of years from 1952 to 2020 is \( 2020 - 1952 = 68 \). The value in 2020 is:
\[
V = 94000 \times (1.073)^{68}
\]
Using a calculator:
\[
(1.073)^{68} \approx 104.67
\]
\[
V \approx 94000 \times 104.67 \approx 9846580 \text{ dollars}
\]

Final Answer for Problem 9:

\[
\boxed{9846580}
\]

---

Problem 10: Population of Bolton

The population of Bolton in 2000 was 11,171. The estimated population growth is 2.1% every 4 years.

#### a) In 2020
- The number of 4-year periods from 2000 to 2020 is \( \frac{2020 - 2000}{4} = 5 \). The population in 2020 is:
\[
P = 11171 \times (1.021)^5
\]
Using a calculator:
\[
(1.021)^5 \approx 1.110
\]
\[
P \approx 11171 \times 1.110 \approx 12400 \text{ people}
\]

#### b) In 2001
- The number of 4-year periods from 2000 to 2001 is \( \frac{2001 - 2000}{4} = 0.25 \). The population in 2001 is:
\[
P = 11171 \times (1.021)^{0.25}
\]
Using a calculator:
\[
(1.021)^{0.25} \approx 1.005
\]
\[
P \approx 11171 \times 1.005 \approx 11228 \text{ people}
\]

#### c) In 2011
- The number of 4-year periods from 2000 to 2011 is \( \frac{2011 - 2000}{4} = 2.75 \). The population in 2011 is:
\[
P = 11171 \times (1.021)^{2.75}
\]
Using a calculator:
\[
(1.021)^{2.75} \approx 1.068
\]
\[
P \approx 11171 \times 1.068 \approx 11950 \text{ people}
\]

#### d) At the beginning of the next century (2100)
- The number of 4-year periods from 2000 to 2100 is \( \frac{2100 - 2000}{4} = 25 \). The population in 2100 is:
\[
P = 11171 \times (1.021)^{25}
\]
Using a calculator:
\[
(1.021)^{25} \approx 1.656
\]
\[
P \approx 11171 \times 1.656 \approx 18540 \text{ people}
\]

Final Answer for Problem 10:

\[
\boxed{12400, 11228, 11950, 18540}
\]

---

Problem 11: Illiteracy Rate

The illiteracy rate in 2017 was 4,324,000 people. The country hopes to cut this number by 11% every 3 years.

#### a) In 2035
- The number of 3-year periods from 2017 to 2035 is \( \frac{2035 - 2017}{3} = 6 \). The illiteracy rate in 2035 is:
\[
I = 4324000 \times (0.89)^6
\]
Using a calculator:
\[
(0.89)^6 \approx 0.558
\]
\[
I \approx 4324000 \times 0.558 \approx 2416000 \text{ people}
\]

#### b) In 2092
- The number of 3-year periods from 2017 to 2092 is \( \frac{2092 - 2017}{3} = 25 \). The illiteracy rate in 2092 is:
\[
I = 4324000 \times (0.89)^{25}
\]
Using a calculator:
\[
(0.89)^{25} \approx 0.042
\]
\[
I \approx 4324000 \times 0.042 \approx 181600 \text{ people}
\]

#### c) In 2156
- The number of 3-year periods from 2017 to 2156 is \( \frac{2156 - 2017}{3} = 46 \). The illiteracy rate in 2156 is:
\[
I = 4324000 \times (0.89)^{46}
\]
Using a calculator:
\[
(0.89)^{46} \approx 0.000001
\]
\[
I \approx 4324000 \times 0.000001 \approx 4 \text{ people}
\]

Final Answer for Problem 11:

\[
\boxed{2416000, 181600, 4}
\]