Q: A savings account offers a 6% annual interest rate, compounded monthly. If you deposit $$\2$$,000$, what will be the balance after 5 years? (Assume no withdrawals.)

$$A = 2000 \left(1 + \frac{0.06}{12}\$$right)^{12 \times 5}$$ \approx \2$$,689$

Q: How much of a 1-gram sample of radium would remain after 100 years? (Half-life = 1600 years)

$$A(100) = 1 \times e$$^{-\frac{\ln 2}$${1600} \times 100} \approx 0.957$$ grams

Q: A family releases 2 rabbits into the wild. The population grows exponentially as $$P = 2 e$$^{0.693 t}$$, where t$ is in months. How many rabbits will there be after 12 months?

$$P = 2 e$$^{0.693 \times 12}$$ \approx 325,510$$

Q: A sample of rabbits grows exponentially according to $$A(t) = 100 \times 0.5^t. $$How many rabbits will there be after 1.5 years?

$$A(1.5) = 100 \times 0.5$$^{1.5}$$ \approx 35.36$$ rabbits

Q: A radioactive sample has a half-life of 1600 years. How much of a 1-gram sample remains after 1000 years?

$$A(1000) = 1 \times e$$^{-\frac{\ln 2}$${1600} \times 1000} \approx 0.65$$ grams

Question 1

Which formula correctly represents the amount $$A$$ after $$t$$ years when a principal $$P is $$invested at an annual interest rate $$r$$, compounded $$n$$ times per year?

Accepted Answer

$$A = P \left(1 + \frac{r}{n}\$$right)^{nt}$$

Question 2

A savings account offers a 6% annual interest rate, compounded monthly. If you deposit $$\2$$,000$, what will be the balance after 5 years? (Assume no withdrawals.)

Accepted Answer

$$A = 2000 \left(1 + \frac{0.06}{12}\$$right)^{12 	imes 5}$$ \approx \2$$,689$

Question 3

Which equation describes the amount $$A(t) of a $$substance remaining after $$t$$ years if it decays exponentially with decay constant $$k$$?

Accepted Answer

$$A(t) = A_0$ e$$^{-kt}$$

Question 4

The half-life of radium is 1600 years. What is the value of $$k in $$the decay formula $$A(t) = A_0$ e$$^{-kt}$$?

Accepted Answer

$$k = \frac{\ln 2}{1600}$$

Question 5

How much of a 1-gram sample of radium would remain after 100 years? (Half-life = 1600 years)

Accepted Answer

$$A(100) = 1 	imes e$$^{-\frac{\ln 2}$${1600} 	imes 100} \approx 0.957$$ grams

Question 6

If an investment is compounded continuously at a rate $$r$$, which formula gives the amount after $$t$$ years?

Accepted Answer

$$A = P e$$^{rt}$$

Question 7

How long does it take for an investment to double if it is compounded continuously at 7% per year?

Accepted Answer

$$t = \frac{\ln 2}{0.07} \approx 9.9$$ years

Question 8

A population of fish in a lake grows according to $$P(t) = 1000 	imes 2$$^{t/3}$$, where t$ is in years. What is the doubling time for this population?

Accepted Answer

$$t = 3$$ years

Question 9

A family releases 2 rabbits into the wild. The population grows exponentially as $$P = 2 e$$^{0.693 t}$$, where t$ is in months. How many rabbits will there be after 12 months?

Accepted Answer

$$P = 2 e$$^{0.693 	imes 12}$$ \approx 325,510$$

Question 10

If the population of fish in a lake is modeled by $$P(t) = 1000 	imes 2$$^{t/3}$$, how long will it take for the population to reach 8000 fish?

Accepted Answer

$$t = 9$$ years

Question 11

A sample of rabbits grows exponentially according to $$A(t) = 100 	imes 0.5^t. $$How many rabbits will there be after 1.5 years?

Accepted Answer

$$A(1.5) = 100 	imes 0.5$$^{1.5}$$ \approx 35.36$$ rabbits

Question 12

If the half-life of a radioactive isotope is $$T_{1/2}$$, what is the general formula for the decay constant $$k$$?

Accepted Answer

$$k = \frac{\ln 2}{T_{1/2}}$$

Question 13

A bank offers 4% annual interest, compounded quarterly. What is the effective annual rate (EAR)?

Accepted Answer

$$EAR = (1 + \frac{0.04}{4})^4$ - 1 \approx 0.0406 or$$ $$4.06\%$$

Question 14

If you want to accumulate $$\11$$,000$ in three years by investing a lump sum at 12% annual interest compounded monthly, how much should you invest now?

Accepted Answer

$$P = \frac{11,000}{(1 + \frac{0.12}{12}$$)^{36}$$} \approx \7$$,800$

Question 15

A radioactive sample has a half-life of 1600 years. How much of a 1-gram sample remains after 1000 years?

Accepted Answer

$$A(1000) = 1 	imes e$$^{-\frac{\ln 2}$${1600} 	imes 1000} \approx 0.65$$ grams

Question 16

Which of the following best describes the difference between compounding interest annually and continuously?

Accepted Answer

Continuous compounding uses $$A = P e$$^{rt}$$, while annual compounding uses A = P (1 + r)^t$

Question 17

A population of bacteria doubles every 4 hours. If the initial population is 500, what will the population be after 12 hours?

Accepted Answer

$$P = 500 	imes 2$$^{12/4}$$ = 500 	imes 2^3$ = 500 	imes 8 = 4,000$$

Question 18

If an investment grows according to $$A = P (1 + r)^t$$, what is the value of $$A$$ after 10 years for $$P = \1$$,000$$ and r = 0.05$?

Accepted Answer

$$A = 1000 	imes (1 + 0.05$$)^{10}$$ \approx \1$$,629$

Exponential Growth, Compound Interest, and Radioactive Decay: Mathematical Models in Chemistry

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