+180 The formula for an investment worth with interest compounded annually is A = P(1+i)^n, where P represents the initial investment, i is the interest rate, and A is the worth of the investment after n years.
a) Rearrange the formula for P. What was the initial investment of an investment worth $1000 that compounded 10% interest for 10 years?
b) Rearrange the formula for i. What is the interest rate of an investment whose worth went from $1000 to $1200 in 2 years?
c) Explain a method with which you could estimate how many years it would take for an investment to reach a certain worth at a certain interest rate.
d) Estimate how many years would it take an investment of $2100 at 20% interest to reach a worth of $5225?
a) P = A (1 + i)^(-n)
P ==1000(1.10)^-10
P ==1000 x 0.3855432
P ==$385.54 - initial investment.
b) i = (A/P)^(1/n) - 1
i ==(1200 / 1000)^(1/2) - 1
i ==Sqrt(1.2) - 1
i ==1.095445 - 1 x 100 ==9.5445% - interest rate.
c) Use "Rule of 72". $100 at 10% will double to $200 in: 72 / 10 [interest rate] ==~7.2 years.
d) 5225 ==2100 x [1 + 0.20]^N. Note: We don't have to "estimate" but will solve it exactly!!!
5225 / 2100 ==1.20^N
2.488095238 ==1.20^N
Take the log of both sides
N ==log(2.488095238) / log(1.20)
N ==5 years for 2100 to grow to 5225 at 20% comp. annually.
