Question

if $12000 is invested at an interest rate of 10% per year compounded monthly. given a(t)=p(1+r over n)^nt. find the amount of the investment at the end of 3 years, and how long will the investment take to double?

Answer 1

a(t)=p( 1+ (r/n) )^(nt)
a(3)=12000(1+ (0.10/12) )^(3*12)

Answer 2

would u be able to tell me what the final answer is because im not sure if i did the equation right

Answer 3

whad did you get for the value?

Answer 4

a(3)=12000(1+ (0.10/12) )^(3*12)
a(3)=12000(1+ (0.10/12) )^(36)
a(3)=12000(1+ 0.00833 )^(36)
a(3)=12000(1.00833)^(36)
a(3)=12000(1.348)

Answer 5

-0.001835789

Answer 6

i see what i did wrong now it makes sense

Answer 7

:D

Answer 8

how long would it take the investment to double? do you know what the formula set up is for that?

Answer 9

ya,
so we are looking for time, or t,
rather than before where we were solving for a(t).
so a(t) = double 12000
or
a(t) = 24000
24000=12000(1+ (0.10/12) )^(t*12)

Answer 10

so the answer is t = 20.9 years correct?

Answer 11

t = 6.9
:(

Answer 12

24000=12000(1+ (0.10/12) )^(t*12)
24000=12000(1.00833 )^(t*12)
24000/12000 = 1.00833^(12t)
2 = 1.00833^(12t)
log(2) = log( 1.00833^(12t))
log(2) = 12t*log( 1.00833)
log(2) / log(1.00833) = 12t
(log(2) / log(1.00833) )/12= t