Asked by Seth
Suppose that $200 was deposited on 1st Jan 2000 into an account that earned 5% interest compounded semiannually. Suppose further that $200 was deposited on 1st Jan 2001 into a different account that earned 6% interest compounded semiannually. In what month of what year will the total amount in the account earning 6% interest overtake the total amount in the account earning 5% interest?
Answers
Answered by
Reiny
so we are solving ......
200( 1.025)^(t) = 200(1.03)^(t-2) , where t is number of half-years
1.025^(t) = 1.03^(t-2)
log both sides
log (1.025^(t)) = log(1.03^(t-2))
t log 1.025 = (t-2) log 1.03
t log 1.025 = t log 1.03 - 2log 1.03
t log 1.025 - t log 1.03 = -2log 1.03
t(log 1.025 - log 1.03) = -2log 1.03
t = -2log 1.03/(log 1.025 - log 1.03)
t = 12.1486 half years after jan 1, 2000
I will leave it to you to figure out the month and year
Remember t is in half years,
so it will take 6 years and .1486(6) months .....
200( 1.025)^(t) = 200(1.03)^(t-2) , where t is number of half-years
1.025^(t) = 1.03^(t-2)
log both sides
log (1.025^(t)) = log(1.03^(t-2))
t log 1.025 = (t-2) log 1.03
t log 1.025 = t log 1.03 - 2log 1.03
t log 1.025 - t log 1.03 = -2log 1.03
t(log 1.025 - log 1.03) = -2log 1.03
t = -2log 1.03/(log 1.025 - log 1.03)
t = 12.1486 half years after jan 1, 2000
I will leave it to you to figure out the month and year
Remember t is in half years,
so it will take 6 years and .1486(6) months .....
Answered by
jason
200(1.025)^((n+1)*2) = 200(1.03)^(2n)
2(n+1)ln1.025 = 2nln1.03
(n+1)ln1.025 = nln1.03
n(ln 1.03 - ln1.025) = ln1.025
n = 5.074 years
year 2006, month January
2(n+1)ln1.025 = 2nln1.03
(n+1)ln1.025 = nln1.03
n(ln 1.03 - ln1.025) = ln1.025
n = 5.074 years
year 2006, month January
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