I Can't seem to get the right answer for this question, please help!
Initial Investment: $1000
Annual Rate: 12%
Time To Double: ?
Use A=P(1+(r/n))^(nt)
Show steps.
2007-06-11 17:48:44 · 2 answers · asked by I have 32 characters 2 work with 3 in Science & Mathematics ➔ Mathematics
Answer is supposed to be 5.78; I keep getting 6.12
2007-06-11 17:53:59 · update #1
Answer is supposed to be 5.78; I keep getting 6.12
2007-06-11 17:54:17 · update #2
Answer is supposed to be 5.78; I keep getting 6.12
2007-06-11 17:54:19 · update #3
Thanks, Sean H!
2007-06-11 18:11:28 · update #4
When I worked the problem using the formula you gave I got 6.12 like you. However, if you assume continuous compounding, then you need to find t so that
e^{t*0.12} = 2
t = ln(2)/0.12 = 5.78
2007-06-11 18:05:03 · answer #1 · answered by Sean H 5 · 1⤊ 0⤋
What you leave out of your information is the value of n, the number of times the interest is compounded each year.
If just once, solve 2 = 1(1.12)^t
ln(2) = t ln(1.12)
t = ln(2)/ln(1.12) = 6.116 years
But to get 5.8 years, n must be 12, compounding monthly:
2 = 1( 1 + 0.12/12)^(12t)
2 = 1.01^(12t)
12t = ln(2)/ln(1.01)
12t = 69.6607 compounding periods
t = 5.8051 years
2007-06-11 18:05:06 · answer #2 · answered by Philo 7 · 0⤊ 1⤋