Compound Interest

Question

Suppose that
$1900

is invested at an interest rate of
3.75%

per year, compounded continuously. After how many years will the initial investment be doubled?

Answer 1

Continuous interest:  A = pe^rt

A = future amount = 2p

e = Euhler's number

r = rate = .0375

t = time in years

 

2p = pe^rt

2 = e^rt

2 = e^.0375t

Take natural log of both sides

 

ln2 = ln(e^.0375t)

ln2 = t[ln e^.0375]

ln2 = .0375t

 

t = ln2/ln.0375 ≅ 18.5 years

Answer 2

2 = e^0.0375t

 

Can you solve for t and answer?