SOLUTION: I have invested $1200 in a bank acount with an interest rate 4.5% compounded monthly. After how many years will my account pass $2100
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Question 1197706: I have invested $1200 in a bank acount with an interest rate 4.5% compounded monthly. After how many years will my account pass $2100 Found 2 solutions by Theo, MathTherapy:Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period.
1 + r is the growth factor per time period.
n is the number of time periods.
in your problem, the equation become:
2100 = 1200 * (1 + .045/12) ^ n
this becomes 2100 = 1200 * 1.00375 ^ n
divide both sides of the equation by 1200 to get:
2100/1200 = 1.00375 ^ n
this becomes 1.75 = 1.00375 ^ n
take the log of both sides of the equation to get:
log(1.75) = n * log(1.00375)
solve for n to get:
n = log(1.75) / log(1.00375) = 149.5105101 months.
confirm by replacing n with that to get:
f = 1200 * 1.00375 ^ 149.5105101 = 2100.
this confirms the value of n is correct.
149.5105101 / 12 = 12.45920918 years.
the account will pass 2100 some time in the 13th year.
You can put this solution on YOUR website!
I have invested $1200 in a bank acount with an interest rate 4.5% compounded monthly. After how many years will my account pass $2100
Use the formula for the FUTURE VALUE of $1:
For this problem: ----- Substituting 1,200 for P, .045 for i, 12 for m, and 2,100 for A/FV
------ Converting to LOGARITHMIC form
Time, or
