Asked by Mea
An investment grows ar a rate of 2.5% each year compounded continuously. Approximately how long will it take the investment to double?
Answers
Answered by
Damon
dx/dt = .025 x
dx/x = .025 dt
ln x = .025 t + c
x = e^(.025 t+c) = e^c e^.025 t
so x = C e^.025 t
at t= 0, x = C e^0 = C
when is x = 2C ???
2C = C e^.025 x
e^.025 x = 2
.025 x = ln 2 = .6931
x = 27.7 years
dx/x = .025 dt
ln x = .025 t + c
x = e^(.025 t+c) = e^c e^.025 t
so x = C e^.025 t
at t= 0, x = C e^0 = C
when is x = 2C ???
2C = C e^.025 x
e^.025 x = 2
.025 x = ln 2 = .6931
x = 27.7 years