You invest $3000 in a savings account where you earn 4% interest compounded continuously

Question

a) what is the formula A(t) for the balance after t years?

A(t)=3000e^0.04(t)

b) How long will it take for the money to double?

6000=3000e^0.04t Is this the correct setup so far or would I just plug in 2 for t?

Answer 1

no, don't sub in t=2

you already took care of the "doubling" when you changed the 3000 to 6000
so you have to solve

6000 = 3000e^(.04t) for t

divide both sides by 3000,
2 = e^(.04t)
take ln of both sides
ln 2 = ln (e^(.04t))
ln 2 = .04t (ln e), but lne = 1
so
.04t = ln 2
t = ln 2/.04 = 17.33 years

Answer 2

dx/dt = .04x
ln x = .04 t + C
x = e^(.04t + C) = ce^(.04 t)
c = 3000
so
x = 3000 e^(.04 t) check
6000/3000 = 2 ON THE LEFT
2 = e^(.04 t)
ln 2 = .04 t
etc