you probably meant
A = P e^(kt) at the end of your post
initial account value was 2500 ---> t = 0 and A = 2500
2500 = P e^0, ---> P = 2500
when t=3, A = 4200
4200 = 2500 e^(3k)
1.68 = e^(3k)
ln both sides
ln 1.68 = ln (e^(3k))
ln 1.68 = 3k(lne), but lne = 1
3k = ln1.68
k = ln 1.68/3
A = 2500 e^(t ln1.68/3)
The rate of change of an investment account earning continuous compound interest is given by
dA/dt=kA
where k is a positive constant. The initial account value was $2500. At the end of the third year, the account value was $4200.
Find the particular solution to the differential equation. You may also use A = Pekt.
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