Question 962515: A person invests 6000 dollars in a bank. The bank pays 6% interest compounded daily.
To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 11100 dollars?
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A person invests 6000 dollars in a bank. The bank pays 6% interest compounded daily. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 11100 dollars?
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A=P(1+r/n)^nt, P=initial investment, r=interest rate, n=number of compounding periods per year
A=amt after t-years
..
assume 365 days/yr
11100=6000(1+.06/365)^365t
11100/6000=(1+.0001644)^365t
111/60=(1.0001644)^365t
take log of both sides
log(111/60)=365tlog(1.0001544)
t=log(111/60)/365log(1.0001544)=14.2
how long must the person leave the money in the bank? about 14 years
Answer by ikleyn(53886)  (Show Source):
You can put this solution on YOUR website! .
person invests 6000 dollars in a bank. The bank pays 6% interest compounded daily.
To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 11100 dollars?
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The solution and the answer in the post by @lwsshak3 both are incorrect due to arithmetic error on the way.
Find my correct solution below.
A=P(1+r/n)^nt, P=initial investment, r=interest rate, n=number of compounding periods per year
A=amt after t-years
..
assume 365 days/yr
11100 = 6000(1+.06/365)^365t
111/60 = (1+0.06/365)^365t
111/60 = (1+0.06/365)^365t
take log of both sides
log(111/60) = 365t*log(1+0.06/365)
t=log(111/60)/(365*log(1+0.06/365)) = 10.25391
how long must the person leave the money in the bank? about 10.25 years. (rounded as requested)
CHECK.  = 11,097 dollars <<<--->>> very close to $11,000.
Solved correctly.
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