Complete the table for a savings account in which interest is compounded continuously. (Round your answers to four decimal places.)

Question

Initial Investment: $350

Annual % Rate: ______ (%)

Time to Double: ______ (yr)

Amount after 13 years: $435.21

Answer 1

For interest compounded continuously, we use:

A(t) = A₀e^rt where A(t) is the amount of money you have at ant time "t", A₀ is the initial amount of money (money at time t = 0), r is the interest rate (usually annual), and t is the time (usually in years).

So if starting with $350 and ending with $435.21 with t = 13 we get:

435.21 = 350e^13r

435.21/350 = e^13r

ln(435.21/350) = ln(e^13r)

ln(435.21/350) = 13r(ln(e))

ln(435.21/350) = 13r(1)

r = ln(435.21/350)/13

r = 0.016761 or 1.6761%

To find the time to double your money using that interest rate:

2A₀ = A₀e^rt

2A₀/A₀ = e^rt

2 = e^rt

ln(2) = ln(e^rt)

ln(2) = (rt)ln(e)

ln(2) = (rt)(1)

t = ln(2)/r

t = ln(2)/(ln(435.21/350)/13) = 41.3543 years