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

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

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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