Compound Interest Formula Explained, Investment, Monthly & Continuously, Word Problems, Algebra

Introduction to Compound Interest Formulas00:00:01

This section introduces two key compound interest formulas. The first, A = P(1 + r/n)^(nt), is used for interest compounded at discrete intervals (monthly, weekly, daily, quarterly, semi-annually, annually). A represents the future value, P the principal (present value), r the annual interest rate (as a decimal), n the number of times interest is compounded per year, and t the time in years. The second formula, A = Pe^(rt), is used specifically when interest is compounded continuously. Here, 'e' is Euler's number (the inverse of the natural log function).

Calculating Future Value with Monthly Compound Interest00:02:49

The first problem demonstrates calculating the future value of an investment. Susan deposits $20,000 at an 8% annual interest rate compounded monthly for 40 years. Using the formula A = P(1 + r/n)^(nt), with P=$20,000, r=0.08, n=12, and t=40, the future value is calculated to be approximately $485,046.79. This highlights the power of saving early.

Determining Present Value for a Future Goal00:04:56

This example focuses on finding the principal amount (P) needed to reach a future financial goal. John wants $2 million in 45 years with a 9.5% annual interest rate compounded quarterly. Given A=$2,000,000, r=0.095, n=4, and t=45, the calculation shows he needs to deposit approximately $29,249.96 today to achieve his goal.

Solving for Interest Rate (R) with Annual Compounding00:07:48

Sarah aims to turn her $10,000 investment into $100,000 in 20 years, compounded annually. The problem involves solving for 'r'. With A=$100,000, P=$10,000, n=1, and t=20, the required annual interest rate is found to be approximately 12.2%.

Calculating Time (T) to Reach a Financial Goal00:10:40

Mary invests $50,000 at an 8.4% annual interest rate compounded semi-annually and wants to reach $1 million. This section demonstrates how to solve for 't' (time). Using logarithms, the calculation shows it will take approximately 36.4 years for her investment to grow to $1 million.

Future Value with Continuously Compounded Interest00:14:31

Juliet invests $100,000 at 7.2% interest compounded continuously for 30 years. Using the formula A = Pe^(rt), with P=$100,000, r=0.072, and t=30, her investment will be worth approximately $867,013.77.

Present Value with Continuously Compounded Interest00:16:11

Mark desires $1.5 million in 50 years with a 12% interest rate compounded continuously. He needs to determine his initial investment (P). With A=$1,500,000, r=0.12, and t=50, the required principal is found to be a surprisingly small amount of approximately $3,718.13 due to the long time horizon and high interest rate.

Solving for Time (T) with Continuously Compounded Interest00:18:18

John invests $5,000 at an 11% interest rate compounded continuously and wants to reach $2 million. This final example demonstrates solving for 't' using natural logarithms. It's calculated that it will take approximately 54.47 years for his investment to reach $2 million, emphasizing the benefit of early and long-term investment.