Question 962515
.
 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)


<U>CHECK</U>.  &nbsp;&nbsp;{{{6000*(1+0.06/365)^(365*10.25)}}} = 11,097 dollars &nbsp;&nbsp;&nbsp;&nbsp;<<<--->>> &nbsp;&nbsp;&nbsp;&nbsp;very close to &nbsp;$11,000.



Solved correctly.