Q: Suppose $$1000 is $$invested at an annual interest rate of $$6\%$$ compounded monthly for $$10$$ years. What is the accumulated value?

$$1000 \times (1 + \frac{0.06}{12}$$)^{120}$$ \approx 1819.40$$

Q: A radioactive material decays according to $$m(t) = 400e$$^{-0.02t}$$. How much of the material remains after 35$ years?

$$400e^{-0.02 \times 35}$$ \approx $$400e^{-0.7}$$ \approx 198.05$$

Question 1

Which of the following is the correct formula for the accumulated value $$P(t) of an $$initial investment $$P_0$$ compounded $$n$$ times per year at an annual interest rate $$r$$ for $$t$$ years?

Accepted Answer

$$P(t) = P_0$ \left(1 + \frac{r}{n} \$$right)^{nt}$$

Question 2

Suppose $$1000 is $$invested at an annual interest rate of $$6\%$$ compounded monthly for $$10$$ years. What is the accumulated value?

Accepted Answer

$$1000 	imes (1 + \frac{0.06}{12}$$)^{120}$$ \approx 1819.40$$

Question 3

A radioactive material decays according to $$m(t) = 400e$$^{-0.02t}$$. How much of the material remains after 35$ years?

Accepted Answer

$$400e^{-0.02 	imes 35}$$ \approx $$400e^{-0.7}$$ \approx 198.05$$

Question 4

If a population of bacteria doubles every $$3$$ hours, which exponential model best represents the population $$y$$ after $$t$$ hours, starting with $$100$$ bacteria?

Accepted Answer

$$y = 100 \cdot 2$$^{t/3}$$

Question 5

Which transformation occurs when the graph of $$f(x) = \log_5$(x) is $$changed to $$f(x) = -5\log_5$(x+4)$$?

Accepted Answer

Shifted left by $$4$$, reflected over $$x-$$axis, and stretched vertically by $$5$$

Question 6

What is the $$x-$$intercept of the logarithmic function $$f(x) = \log_5$(x-4)$$?

Accepted Answer

Question 7

If $$P_0 = 1000$, r = 0.06$, and n = 8760$ (hourly compounding), what is the accumulated value after 10$ years?

Accepted Answer

$$1000 \left(1 + \frac{0.06}{8760} \$$right)^{87600}$$ \approx 1822.12$$

Question 8

Which formula is used for continuous compounding of interest?

Accepted Answer

$$P(t) = P_0$ e$$^{rt}$$

Question 9

How long will it take for an investment to double if compounded continuously at $$8\%$$ annual interest?

Accepted Answer

$$t = \frac{\ln 2}{0.08} \approx 8.66$$ years

Question 10

A school’s enrollment grows exponentially. If there were $$1200$$ students in $$2006$$ and $$1500 in$$ $$2010$$, what is the exponential growth model for the enrollment $$E(t)$$, where $$t is $$years since $$2006$$?

Accepted Answer

$$E(t) = 1200 \cdot (1.062)^t$$

Question 11

The minimum wage in $$1990$$ was $$3.80$$ and in $$2000$$ was $$5.15. $$Assuming exponential growth, what is the model $$f(t)$$ for the wage, where $$t is $$years since $$1990$$?

Accepted Answer

$$f(t) = 3.80 \cdot (1.030)^t$$

Question 12

A certain type of bacteria starts with $$100$$ and doubles every $$3$$ hours. How long will it take for the population to reach $$200,000$$?

Accepted Answer

$$t = 3 \cdot \frac{\ln(2000)}{\ln(2)} \approx 32.9$$ hours

Question 13

If $$P_0 = 1000$, r = 0.06$, and n = 525$$,600$$ (compounded every minute), what is the accumulated value after 10$ years?

Accepted Answer

$$1000 \left(1 + \frac{0.06}{525,600} \$$right)^{5,256,000}$$ \approx 1822.12$$

Question 14

A painting is appraised at $$14$$ million dollars in $$2020$$ and is growing in value exponentially at $$5\%$$ per year. What will its value be in $$2030$$?

Accepted Answer

$$14 \cdot (1.05$$)^{10}$$ \approx 22.77$$ million

Question 15

If a substance decays exponentially and $$270$$ grams are present after $$6$$ hours, and $$242$$ grams after $$10$$ hours, what is the decay rate $$b in $$the model $$y = a b^t$$?

Accepted Answer

$$b = \left(\frac{242}{270}\$$right)^{1/4}$$ \approx 0.979$$

Question 16

What is the formula for the half-life $$t of a $$radioactive material decaying exponentially as $$m(t) = m_0$ e$$^{-kt}$$?

Accepted Answer

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

Question 17

If $$P_0 = 1000$, r = 0.06$, and t = 10$, what is the accumulated value if compounded continuously?

Accepted Answer

$$1000e^{0.6}$$ \approx 1822.12$$

Question 18

A population grows according to $$P(t) = 5000e$$^{0.03t}$$. How long will it take for the population to reach 10$$,000$$?

Accepted Answer

$$t = \frac{\ln(2)}{0.03} \approx 23.1$$ years

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