Find the time it takes for $9,100 to double when invested at an annual interest rate of 3%, compounded continuously.

Answer 1

2 = e^.03t

take natural logs of both sides

ln2 = .03t

t = ln2/.03

t =about 23.1 years to double the investment of $9100 to $18,200, or the time for any investment of any amount to double

general formula for continuous compoundings is

A=Pe^rt where r= annual interest rate, t= number of years, P=original amount invested, A= Amount after t years, at r annual interest, e = the base of the natural log