Serenity deposits $1, 000 every year into an account earning an annual interest rate of 6.5%, compounded annually. How many years would it be until Serenity had $25, 000 in the account, to the nearest tenth of a year? Use the following formula to determine your answer.
A = d(((1 + i) ^ n - 1)/i)
A = the future value of the account after n periods
d = the amount invested at the end of each period
i = the interest rate per period
n = the number of period
Let's plug in the given values into the formula:
A = $25,000
d = $1,000
i = 6.5% = 0.065
n = number of periods
25000 = 1000(((1 + 0.065) ^ n - 1)/0.065)
Simplify the equation:
25 = ((1.065) ^ n - 1)/0.065
25 * 0.065 = (1.065) ^ n - 1
1.625 = (1.065) ^ n - 1
2.625 = (1.065) ^ n
Take the natural log of both sides:
ln(2.625) = ln((1.065) ^ n)
ln(2.625) = n * ln(1.065)
n = ln(2.625) / ln(1.065)
n ≈ 25.7
It would be approximately 25.7 years until Serenity had $25,000 in the account.