Continuous Compound Interest Word Problems Lesson

In the last lesson, we learned how to solve a word problem that involved the compound interest formula. For a given time period, as we increase the number of compounding periods or how often interest is deposited in our account, the account balance gets larger, but only to a certain point. There is a maximum amount of interest that you can earn by increasing the number of compounding periods for a given time period and a given interest rate.
Note: The special number e used in the formula is known as Euler's number (pronounced as Oiler's number). This number comes up quite a bit in mathematics. The approximate value for this number is about 2.718. It is the base of the Natural Logarithm, that we will learn about in a few lessons. Let's look at the formula for continuous compound interest.

Continuous Compound Interest Formula

$$A=Pe^{rt}$$

• A is the account balance
• P is the amount invested or the principal
• e is a special number known as Euler's number
• r is the rate as a decimal
• t is the time in years

Let's look at an example.
Example #1: Solve each word problem. Round your answer to the nearest hundredth.
Mia invests $1,989 in a savings account with a fixed annual interest rate of 3% compounded continuously. What will the account balance be after 8 years?
We only need to plug into our formula. $$A=\hspace{.1em}?$$ $$P=1{,}989$$ $$r=0.03$$ $$t=8$$ $$A=Pe^{rt}$$ $$2{,}528.51=1{,}989e^{0.03 \cdot 8}$$ After 8 years, the account balance is about $2,528.51.
Example #2: Solve each word problem. Round your answer to the nearest hundredth.
Jacob invests a sum of money in a savings account with a fixed annual interest rate of 5% compounded continuously. After 15 years, the balance reaches $4,621.41. What was the amount of the initial investment?
We only need to plug into our formula. $$A=4{,}621.41$$ $$P=\hspace{.1em}?$$ $$r=0.05$$ $$t=15$$ $$A=Pe^{rt}$$ Solve for P: $$P=\frac{A}{e^{rt}}$$ $$P=\frac{4{,}621.41}{e^{0.05 \cdot 15}}$$ $$P=2{,}183$$ The inital amount invested was about $2,183.
Example #3: Solve each word problem. Round your answer to the nearest hundredth.
Kristin invests $2,239 in a 401(k) retirement account with a fixed annual interest rate compounded continuously. After 16 years, the balance reaches $6,862.21. What is the interest rate of the account?
We only need to plug into our formula. $$A=6{,}862.21$$ $$P=2{,}239$$ $$r=\hspace{.1em}?$$ $$t=16$$ $$A=Pe^{rt}$$ Solve for r:
At this point, we haven't covered logarithms in our course but most of you have taken Algebra 2, where this topic is covered. We will just give the basics here and cover it in more detail in the coming lessons.
Divide both sides by P: $$e^{rt}=\frac{A}{P}$$ Take the Natural Log (ln) of each side. This is a logarithm with a base of e, the special number discussed above. $$\text{ln}\left(e^{rt}\right)=\text{ln}\left(\frac{A}{P}\right)$$ From the properties of logarithms, the exponent on e (rt) can come out in front: $$rt \cdot \text{ln}\left(e\right)=\text{ln}\left(\frac{A}{P}\right)$$ By rule ln(e) = 1 $$rt=\text{ln}\left(\frac{A}{P}\right)$$ Divide both sides by t: $$r=\frac{\text{ln}\left(\frac{A}{P}\right)}{t}$$ $$r=\frac{\text{ln}\left(\frac{6{,}862.21}{2{,}183}\right)}{16}$$ $$r=0.07$$ The interest rate on the account is about 7%.

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