To solve this problem, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the amount of money accumulated after n years, including interest
P = the principal amount (the initial amount of money invested)
r = the annual interest rate (in decimal form)
n = the number of times that interest is compounded per year
t = the number of years the money is invested for
In this case, P = $5000, r = 0.10 (10%), n = 4 (quarterly compounding), and we want to find the value of t when A = $7500.
7500 = 5000(1 + 0.10/4)^(4t)
7500 = 5000(1 + 0.025)^(4t)
7500 = 5000(1.025)^(4t)
1.5 = 1.025^(4t)
Using logarithms to solve for t:
log(1.5) = log(1.025^(4t))
log(1.5) = 4t * log(1.025)
log(1.5) / log(1.025) = 4t
t = log(1.5) / (4 * log(1.025))
t ≈ log(1.5) / (4 * 0.0100886)
t ≈ log(1.5) / 0.0403544
t ≈ 3.09 years
Therefore, it will take approximately 3.09 years for the investment to grow to $7500 with a 10% annual interest rate compounded quarterly.
1. Unit 8: A $5000 investment is made in a savings account that pays 10%/a compounded quarterly. How long will it take for the investment to grow to $7500?
1 answer