Given:
d = $1,000
i = 6.5% = 0.065
A = $25,000
Using the formula provided:
$25,000 = $1,000((1 + 0.065)^n - 1) / 0.065
Solving for n:
25 = ((1.065)^n - 1) / 0.065
25 * 0.065 = (1.065)^n - 1
1.625 = (1.065)^n - 1
1.625 + 1 = (1.065)^n
2.625 = (1.065)^n
Taking the natural logarithm 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 ≈ 15.7
Therefore, it would take Serenity approximately 15.7 years to have $25,000 in the account.
Serenity deposits dollar sign, 1, comma, 000$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 dollar sign, 25, comma, 000$25,000 in the account, to the nearest tenth of a year? Use the following formula to determine your answer.
A, equals, d, left bracket, start fraction, left bracket, 1, plus, i, right bracket, to the power n , minus, 1, divided by, i, end fraction, right bracket
A=d(
i
(1+i)
n
−1
)
A, equalsA= the future value of the account after n periods
d, equalsd= the amount invested at the end of each period
i, equalsi= the interest rate per period
n, equalsn= the number of periods
1 answer
1 answer