Gabriella deposits dollar sign, 550$550 every year into an account earning an annual interest rate of 7.9%, compounded annually. How many years would it be until Gabriella 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
)
Given:
A (amount needed): $11,000
d (annual deposit): $550
i (annual interest rate): 7.9% or 0.079
We will use the formula provided to find the number of years needed:
A=d(i(1+i)^n -1) / i
Substitute the given values:
11,000 = 550(0.079(1+0.079)^n - 1) / 0.079
Simplify:
(11,000 / 0.055) + 550 = (550)(1.079)^n
(200,000) + 550 = (594.5)^n
200,550 = 594.5^n
Now, isolate the exponent:
log(200,550) = log(594.5)^n
log(200,550) = n * log(594.5)
n = log(200,550) / log(594.5)
n ≈ 5.5
It would be approximately 5.5 years until Gabriella had $11,000 in the account.