Find the time it takes for $7,400 to double when invested at an annual interest rate of 19%, compounded continuously.

Answer 1

A = Pe^rt where A is $14800, P = $7400, r = 0.19 and t is the unknown.

14800 = 7400e^0.19t Divide by 7400

2 = e^0.19t Take natural log of both sides.

ln 2 = 0.19t (ln e) (ln e = 1)

0.6931 = 0.19t

t = 3.6 years

P.S. There is also the rule of 72 (or 70) to determine estimated doubling time. Divide 72 or 70 by the interest rate.

In this case you get answers of 3.8 and 3.7 years respectively (close!!)

Answer 2

A=Pe^rt

Doubling

A/P = 2 = e^rt Note: Value of P doesn't matter

rt=ln(2)

t = ln(2)/.19 ≈ 3.648 years

≈ 3 years, 7months, 23 days, 7 hours, 57 minutes and 23 seconds

Answer 3

2= e^.19t

ln2 = .19t

t = ln2/.19 = 3.648 years to double any amount of money

= 3 years 236 days and 17 hours 37 minutes 19.29 seconds

that's the time for $7400 to become $14,800

the time for $100,000 to become $200,000

or the time for literally any amount P to become 2P