How much would $100 invested at 8% interest compounded continuously be worth after 15 years? Round your answer to the nearest cent.
[tex]A(t)=P \bullet e^{rt}[/tex]
A. $285.67
B. $332.01
C. $220.00
D. $317.22
Answer (1)
Substitute the given values into the formula for continuous compounding: A ( t ) = P × e r t .
Calculate the exponent: 0.08 × 15 = 1.2 .
Calculate e 1.2 ≈ 3.3201169227365472 .
Multiply the result by 100 and round to the nearest cent: A ( 15 ) ≈ $332.01 .
Explanation
Understanding the Problem We are given an initial investment of $100 at an interest rate of 8% compounded continuously for 15 years. We need to find the final amount. The formula for continuous compounding is given by A ( t ) = P × e r t , where A ( t ) is the amount after time t , P is the principal amount, r is the interest rate, and t is the time in years.
Substituting the Values We are given P = 100 , r = 0.08 , and t = 15 . We need to substitute these values into the formula A ( t ) = P × e r t .
Calculating the Exponent Substituting the values, we get A ( 15 ) = 100 × e 0.08 × 15 .
Calculating the Exponential Term First, we calculate the exponent: 0.08 × 15 = 1.2 .
Calculating the Final Amount Now we need to calculate e 1.2 . From the tool, we have e 1.2 ≈ 3.3201169227365472 .
Rounding to the Nearest Cent Now, we multiply this by 100: A ( 15 ) = 100 × 3.3201169227365472 = 332.01169227365472 .
Final Answer Finally, we round the amount to the nearest cent: A ( 15 ) ≈ 332.01 .
Examples
Continuous compounding is a concept often used in finance to model investments. For example, if you invest $5 , 000 in a fund that offers an annual interest rate of 6% compounded continuously, you can calculate the future value of your investment after a certain period, say 10 years, using the formula A ( t ) = P × e r t . This calculation helps you understand the potential growth of your investment over time and make informed financial decisions. Understanding continuous compounding is also crucial in various fields such as physics and engineering, where exponential growth and decay models are frequently used.
