SOLUTION: suppose that $14018 is invested at an interest rate of 6.3% per year, compounded continuously. Find the exponential function that describes the amount in the account after time t,

Question 1054018: suppose that $14018 is invested at an interest rate of 6.3% per year, compounded continuously.
Find the exponential function that describes the amount in the account after time t, in years.
what is the doubling time?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the continuoous compounding formula is:

f = p * e^(r*n)

f is the future value
p is the present value
e is the scientific constant of 2.718281828
r is the interest rate per time period.
n is the number of time periods

in your problem:

p = 14018
r = .063 per year
n = t years

the formula becomes f = 14018 * e^(.063 * t)

you want to know what is the doubling time.

if 14018 doubles, then it is worth 28036.

the formula becomes 28036 = 14018 * e^(.063 * t)

divide both sides of the equation by 14018 to get 28036 / 14018 = e^(.063 * t)

simplify to get 2 = e^(.063 * t)

take the natural log of both sides of the equation to get ln(2) = ln(e^(.063 * t).

since ln(e^x) = x*ln(e), your equation becomes ln(2) = .063 * t * ln(e).

since ln(e) = 1, your equation becomes ln(2) = .063 * t

divide both sides of the equation by .063 to get ln(2) / .063 = t.

solve for t to get t = ln(2) / .063 = 11.0023362

thqt's the number of years it would take for the money to double at 6.3% per year using continuous compounding.

14018 * e^(.063 * 11.0023362) is equal to 28036.
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