Q: If \$$2000 is invested at an annual interest rate of 3$$\%$ compounded continuously, what will the investment be worth after 8 years? (Round to the nearest dollar.)

$$A = 2000e$$^{0.03 \times 8}$$ \approx \2541$

Question 1

A sum of \$$1000 grows to \1491.82 in 10 $$years under continuous compounding. Using the formula $$A = Pe$$^{rt}$$, what is the interest rate r$ (rounded to the nearest tenth of a percent)?

Accepted Answer

$$r = 0.04 or$$ $$4\%$$

Question 2

Given the formula for continuous compounding $$A = Pe$$^{rt}$$, which of the following correctly isolates r$ in terms of A$, P$, and t$?

Accepted Answer

$$r = \frac{1}{t} \ln\left(\frac{A}{P}\right)$$

Question 3

If \$$2000 is invested at an annual interest rate of 3$$\%$ compounded continuously, what will the investment be worth after 8 years? (Round to the nearest dollar.)

Accepted Answer

$$A = 2000e$$^{0.03 	imes 8}$$ \approx \2541$

Question 4

A scientist deposits \$$5000 in an account that earns interest compounded continuously at a rate of 2.5$$\%$ per year. How many years will it take for the investment to double?

Accepted Answer

$$t = \frac{\ln(2)}{0.025} \approx 27.7$$ years

Question 5

Which of the following best describes the effect of increasing the interest rate $$r in $$the formula $$A = Pe$$^{rt}$$, assuming P$ and t$ are constant?

Accepted Answer

The future value $$A$$ increases exponentially as $$r$$ increases.

Question 6

Suppose \$$1200 is invested at a continuously compounded rate. After 5 years, the investment is worth \1400. $$What is the annual interest rate $$r$$? (Round to the nearest hundredth of a percent.)

Accepted Answer

$$r = \frac{1}{5} \ln\left(\frac{1400}{1200}\right) \approx 0.0312 or$$ $$3.12\%$$

Question 7

A medical research fund needs to grow from \$$10,000 to \15$$,000 in 7 years. What continuous annual interest rate is required? (Round to the nearest tenth of a percent.)

Accepted Answer

$$r = \frac{1}{7} \ln\left(\frac{15000}{10000}\right) \approx 0.0576 or$$ $$5.8\%$$

Question 8

Which of the following statements is true about the formula $$A = Pe$$^{rt}$$?

Accepted Answer

It models exponential growth due to continuous compounding.

Question 9

A company wants to triple its investment in 12 years with continuous compounding. What annual interest rate is needed? (Round to the nearest hundredth of a percent.)

Accepted Answer

$$r = \frac{1}{12} \ln(3) \approx 0.0916 or$$ $$9.16\%$$

Question 10

If \$$500 is invested at a continuous rate of 6$$\%$$ per year, how long will it take to reach \1000$$?

Accepted Answer

$$t = \frac{\ln(2)}{0.06} \approx 11.55$$ years

Question 11

Which of the following is NOT a correct step in solving for $$r in $$the equation $$A = Pe$$^{rt}$$?

Accepted Answer

Multiply both sides by $$e^{-rt}.$$

Question 12

A savings account grows from \$$2500 to \4000 in 9 $$years with continuous compounding. What is the annual interest rate? (Round to the nearest tenth of a percent.)

Accepted Answer

$$r = \frac{1}{9} \ln\left(\frac{4000}{2500}\right) \approx 0.0512 or$$ $$5.1\%$$

Question 13

If \$$800 is invested at a continuous rate of 2$$\%$ per year, what will the balance be after 15 years? (Round to the nearest dollar.)

Accepted Answer

$$A = 800e$$^{0.02 	imes 15}$$ \approx \1082$

Question 14

A hospital endowment must reach \$$50,000 in 20 years. If \20$$,000 is invested now at a continuous rate, what annual interest rate is required? (Round to the nearest hundredth of a percent.)

Accepted Answer

$$r = \frac{1}{20} \ln\left(\frac{50000}{20000}\right) \approx 0.0455 or$$ $$4.55\%$$

Question 15

Which of the following best explains why continuous compounding yields more interest than annual compounding for the same rate and time?

Accepted Answer

Interest is added at every instant, so the principal grows faster.

Exponential and Logarithmic Functions: Continuous Compounding Applications

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