Pedro deposits dollar sign, 700$700 every month into an account earning a monthly interest rate of 0.4%. How many years would it be until Pedro had dollar sign, 28, comma, 000$28,000 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
to solve this problem use this as an example:
Khalil deposits dollar sign, 640$640 every year into an account earning an annual interest rate of 3%, compounded annually. How many years would it be until Khalil had dollar sign, 8, comma, 000$8,000 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
d, equals, 640, left bracket number of dollars per year right bracket
d=640 (number of dollars per year)
A, equals, 8000, left bracket future account value right bracket
A=8000 (future account value)
i, equals, 0, point, 0, 3, left bracket interest rate per year right bracket
i=0.03 (interest rate per year)
Plug everything in:
Plug everything in:
8000, equals, 640, left bracket, start fraction, left bracket, 1, plus, 0, point, 0, 3, right bracket, to the power n , minus, 1, divided by, 0, point, 0, 3, end fraction, right bracket
8000=640(
0.03
(1+0.03)
n
−1
)
8000, equals, 640, left bracket, start fraction, 1, point, 03, to the power n , minus, 1, divided by, 0, point, 0, 3, end fraction, right bracket
8000=640(
0.03
1.03
n
−1
)
start fraction, 8000, divided by, 640, end fraction, equals, start fraction, 640, left bracket, start fraction, 1, point, 03, to the power n , minus, 1, divided by, 0, point, 0, 3, end fraction, right bracket, divided by, 640, end fraction
640
8000
=
640
640(
0.03
1.03
n
−1
)
12, point, 5, equals, start fraction, 1, point, 03, to the power n , minus, 1, divided by, 0, point, 0, 3, end fraction
12.5=
0.03
1.03
n
−1
0, point, 0, 3, left bracket, 12, point, 5, right bracket, equals, left bracket, start fraction, 1, point, 03, to the power n , minus, 1, divided by, 0, point, 0, 3, end fraction, right bracket, left bracket, 0, point, 0, 3, right bracket
0.03(12.5)=
(
0.03
1.03
n
−1
)(0.03)
0, point, 3, 7, 5, equals, 1, point, 03, to the power n , minus, 1
0.375=
1.03
n
−1
plus, 1, plus, 1
+1=
1.03
n
+1
1, point, 3, 7, 5, equals, 1, point, 03, to the power n
1.375=
1.03
n
log, left bracket, 1, point, 3, 7, 5, right bracket, equals, log, left bracket, 1, point, 03, to the power n , right bracket
log(1.375)=
log(1.03
n
)
log, left bracket, 1, point, 3, 7, 5, right bracket, equals, n, log, left bracket, 1, point, 0, 3, right bracket
log(1.375)=
nlog(1.03)
start fraction, log, left bracket, 1, point, 3, 7, 5, right bracket, divided by, log, left bracket, 1, point, 0, 3, right bracket, end fraction, equals, start fraction, n, log, left bracket, 1, point, 0, 3, right bracket, divided by, log, left bracket, 1, point, 0, 3, right bracket, end fraction
log(1.03)
log(1.375)
=
log(1.03)
nlog(1.03)
10, point, 7, 7, 3, 5, 6, 6, 8, 2, equals, n
10.77356682=
n
n, approximately equals, 10, point, 8, years
n≈10.8 years
Khalil would have , dollar sign, 8, comma, 000, after 10.8 years.
Khalil would have $8,000 after 10.8 years.
Using the same formula, we need to plug in the values for Pedro's situation:
d = $700 (amount deposited every month)
A = $28,000 (future account value)
i = 0.004 (monthly interest rate = 0.4%)
n = number of months
$28,000 = $700 * [(1 + 0.004)^n - 1] / 0.004
$28,000 = $700 * [(1.004)^n - 1] / 0.004
$28,000 = $700 * [1.004^n - 1] / 0.004
40 = 1.004^n - 1
41 = 1.004^n
n = log(41) / log(1.004)
n = 96.4 months
Therefore, it would take approximately 96.4 months for Pedro to have $28,000 in the account, which is approximately 8 years.