Compound interest

Question

You invest $1500 into an account which has interest compounded continuously at a fixed rate. After 2 years, the account has $2200. After how many years (from the time of the original investment) will the account have $4000? Give your answer to the nearest hundredth.

Answer 1

Typically, interest compounded continuously is given the formula:

where "A" is the amount of money you have at any time "t", "P" is the initial principal invested, and "r" is the annual interest rate. So, plugging in P = 1500, A = 2200, and t = 2, we can solve for the interest rate "r":

2200 = 1500e^(r)(2)

Divide both sides by 1500 to get:

(22/15) = e^2r

Now, take the natural log of both sides to get:

ln(22/15) = ln(e^2r)

One of the rules of logarithms is that an exponent inside a log can be moved out front as a multiplier:

ln(22/15) = (2r)•ln(e)

A logarithm identity is that ln(e) = 1 so:

ln(22/15) = 2r

Divide both sides by 2 (or multiply both sides by 1/2):

r = (1/2)•ln(22/15)

Plug this in a calculator and you get r = 0.191496

Now, our equation becomes:

A = 1200e^0.191496t

To find out what time "t" is required to get $4000, plug in 4000 and solve for "t":

4000 = 1500e^0.191496t

4000/1500 = e^0.191496t

ln(4000/1500) = ln(e^0.191496t)

ln(4000/1500) = (0.191496t)•ln(e)

ln(4000/1500) = 0.191496t

t = ln(4000/1500)/0.191496

Using your calculator:

t = 5.12 years

Answer 2

2200 = 1500e^2r

e^2r = 22/15

2r = ln(22/15)

r = ln(22/15)/2 = about 19.15%

4000 = 1500e^t(ln(22/15)/2)

e^t(ln(22/15)/2) = 40/15 = 8/3

t(ln(22/15)/2) = ln(8/3

t = ln(8/3)/ln(22/15)/2

t = 2ln(8/3)/ln(22/15) =5.122 years

= about 5 years 1.5 months