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Question 1140562: How long does it take for $1875 to double if it is invested at 10% compounded continuously? Round your answer to two decimal places, if necessary.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the continuous compounding formula is f = p * e ^ (r * t)
f is the future value
p is the present value
e is the scientific constant equal to 2.718281828.....
r is the interest rate per time period
t is the number of time periods
the time period for this problem will be in years.
in this problem, .....
future value is double the present value = 3750
present value is 1875
interest rate per time period is 10% / 100 = .10
number of time periods is what you want to find.
f = p * e ^ (r * t) becomes:
3750 = 1875 * e ^ (.10 * t)
divide both sides of this equation by 1875 to get:
3750/1875 = e ^ (.10 * t)
simplify and take the natural log of both sides of the equation to get:
ln(2) = ln(e ^ (.10 * t)
since ln(e ^ (.10 * t) = .10 * t * ln(e) and since ln(e) = 1, the equation becomes:
ln(2) = .10 * t
solve for t to get t = ln(2) / .10 = 6.931471806 years.
the money will double in 6.931471806 years.
confirm by replacing t with 6.931471806 in the original equation to get:
3750 = 1875 * e ^ (.10 * 6.931471806) becomes 3750 = 3750.
this confirms the solution is correct.
your solution is it will take 6.93 years for the money to double.
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