Question 1140196: solve: suppose that 2000 is invested at interest rate k, compounded continuously, and grows to 2358.79 in 3 yrs. find the exponential growth function.
find the doubling time
please show work
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the continuous compounding formula is f = p * e^(r * n)
f is the future value
p is the present value
r is the interest rate per time period.
n is the number of time periods.
when r = k, the formula becomes f = p * e ^ (k * n)
when f = 2358.79 and p = 2000 and n = 3 years, the formula becomes:
2358.79 = 2000 * e ^ (3 * k)
divide both sides of this equation by 2000 to get:
2358.79 / 2000 = e ^ (3 * k)
take the natural log of both sides of this equation to get:
ln(2358.79 / 2000) = ln(e ^ (3 * k))
since ln(e ^ (3 * k)) is equal to 3 * k * ln(e) and since ln(e) is equal to 1, the equation becomes:
ln(2358.79 / 2000) = 3 * k
divide both sides of this equation by 3 to get:
ln(2358.79 / 2000) / 3 = k
solve for k to get k = .0550005317
your exponential growth function is:
2358.79 = 2000 * e ^ (.0550005317 * 3).
evaluate this function to get 2358.79 = 2358.79, confirming the solution is correct.
to find the doubling time, the formula becomes:
2 = e ^ (.0550005317 * n)
take the natural log of both sides of this equation to get:
ln(2) = ln(e ^ (.0550005317 * n)
this becomes ln(2) = .0550005317 * n
divide both sides by .0550005317 to get:
ln(2) / .0550005317 = n
solve for n to get n = 12.602555418
replace n in the doubling formula to get:
2 = e ^ (.0550005317 * 12.602555418)
evaluate this function to get 2 = 2, confirming the solution is correct.
note that 2 = e ^ (.0550005317 * n) is the same as 2 = 1 * e ^ (.0550005317 * n)
you future value is 2 and your present value is 1, hence the doubling.
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