Gabriel deposits dollar sign, 400$400 every month into an account earning an annual interest rate of 6.3%, compounded monthly. How many years would it be until Gabriel had dollar sign, 11, comma, 000$11,000 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
In this case,
A = $11,000
d = $400
i = 0.063/12 (monthly interest rate)
We need to find n.
$11,000 = $400 * ((0.063/12) * (1 + (0.063/12))^n - 1) / (0.063/12)
Solving for n:
11000 = 400 * ((0.00525) * (1.00525)^n - 1) / 0.00525
11000 = 400 * (1.00525)^n - 400
11000 / 400 = (1.00525)^n - 1
27.5 = (1.00525)^n - 1
28.5 = (1.00525)^n
Taking the natural log of both sides:
ln(28.5) = n * ln(1.00525)
n ≈ ln(28.5) / ln(1.00525)
n ≈ 91.5
So, it would take approximately 91.5 years for Gabriel to have $11,000 in the account.