Compound Interest Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

medium
Find the amount when \2{,}000isinvestedat is invested at 5\%$ compounded continuously for 4 years.

Solution

  1. 1
    Use the continuous compounding formula A=PertA = Pe^{rt} with P=2000P = 2000, r=0.05r = 0.05, t=4t = 4.
  2. 2
    A=2000e0.05ร—4=2000e0.2A = 2000e^{0.05 \times 4} = 2000e^{0.2}.
  3. 3
    e0.2โ‰ˆ1.2214e^{0.2} \approx 1.2214, so Aโ‰ˆ2000ร—1.2214โ‰ˆ2442.81A \approx 2000 \times 1.2214 \approx 2442.81.

Answer

Aโ‰ˆ$2,442.81A \approx \$2{,}442.81
Continuous compounding uses A=PertA = Pe^{rt}, which is the limiting case as compounding frequency nโ†’โˆžn \to \infty. It gives slightly more than any finite compounding frequency.

About Compound Interest

Interest calculated on both the initial principal and the accumulated interest from previous periods. The formula A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt} gives the amount after tt years, and A=PertA = Pe^{rt} gives the continuously compounded amount.

Learn more about Compound Interest โ†’

More Compound Interest Examples

Example 1 easy

You invest [formula]5{,}000[formula]6%$ annual interest compounded quarterly. Find the amount after

Example 2 medium

How long does it take for an investment to double at [formula] annual interest compounded monthly?

Example 4 easy

You deposit [formula]1{,}200[formula]4%$ compounded annually. What is the balance after 5 years?

Example 5 hard

What annual interest rate, compounded monthly, is needed to grow [formula]3{,}000[formula][formula]

โ† All Compound Interest Examples Compound Interest Concept