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
the answer is not 91.5
TO SOLVE here is an example A= 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, 800, left bracket number of dollars per month right bracket
d=800 (number of dollars per month)
A, equals, 10000, left bracket future account value right bracket
A=10000 (future account value)
i, equals, 0, point, 0, 0, 3, 5, left bracket interest rate per month right bracket
i=0.0035 (interest rate per month)
Plug everything in:
Plug everything in:
10000, equals, 800, left bracket, start fraction, left bracket, 1, plus, 0, point, 0, 0, 3, 5, right bracket, to the power n , minus, 1, divided by, 0, point, 0, 0, 3, 5, end fraction, right bracket
10000=800(
0.0035
(1+0.0035)
n
−1
)
10000, equals, 800, left bracket, start fraction, 1, point, 0, 0, 35, to the power n , minus, 1, divided by, 0, point, 0, 0, 3, 5, end fraction, right bracket
10000=800(
0.0035
1.0035
n
−1
)
start fraction, 10000, divided by, 800, end fraction, equals, start fraction, 800, left bracket, start fraction, 1, point, 0, 0, 35, to the power n , minus, 1, divided by, 0, point, 0, 0, 3, 5, end fraction, right bracket, divided by, 800, end fraction
800
10000
=
800
800(
0.0035
1.0035
n
−1
)
12, point, 5, equals, start fraction, 1, point, 0, 0, 35, to the power n , minus, 1, divided by, 0, point, 0, 0, 3, 5, end fraction
12.5=
0.0035
1.0035
n
−1
0, point, 0, 0, 3, 5, left bracket, 12, point, 5, right bracket, equals, left bracket, start fraction, 1, point, 0, 0, 35, to the power n , minus, 1, divided by, 0, point, 0, 0, 3, 5, end fraction, right bracket, left bracket, 0, point, 0, 0, 3, 5, right bracket
0.0035(12.5)=
(
0.0035
1.0035
n
−1
)(0.0035)
0, point, 0, 4, 3, 7, 5, equals, 1, point, 0, 0, 35, to the power n , minus, 1
0.04375=
1.0035
n
−1
plus, 1, plus, 1
+1=
1.0035
n
+1
1, point, 0, 4, 3, 7, 5, equals, 1, point, 0, 0, 35, to the power n
1.04375=
1.0035
n
log, left bracket, 1, point, 0, 4, 3, 7, 5, right bracket, equals, log, left bracket, 1, point, 0, 0, 35, to the power n , right bracket
log(1.04375)=
log(1.0035
n
)
log, left bracket, 1, point, 0, 4, 3, 7, 5, right bracket, equals, n, log, left bracket, 1, point, 0, 0, 3, 5, right bracket
log(1.04375)=
nlog(1.0035)
start fraction, log, left bracket, 1, point, 0, 4, 3, 7, 5, right bracket, divided by, log, left bracket, 1, point, 0, 0, 3, 5, right bracket, end fraction, equals, start fraction, n, log, left bracket, 1, point, 0, 0, 3, 5, right bracket, divided by, log, left bracket, 1, point, 0, 0, 3, 5, right bracket, end fraction
log(1.0035)
log(1.04375)
=
log(1.0035)
nlog(1.0035)
12, point, 2, 5, 5, 6, 8, 2, 4, 4, equals, n
12.25568244=
n
n, approximately equals, 12, months
n≈12 months
William would have , dollar sign, 10, comma, 000, after 12 months.
William would have $10,000 after 12 months.