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

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).

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.

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.

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%.

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.

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.

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.

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.