Question 168828: How long does it take $1000 to double if invested at 7% interest compounded continuously?
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! formula for continuous compounding is:
FV = PV * e^(i*x)
where i is the interest rate
and x is the number of years.
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for your problem:
PV = $1000
FV = $2000 (double the initial investment of $1000)
i = .07
x = what you want to find.
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FV = PV * e^(i*x)
becomes:
2000 = 1000 * e^(.07*x)
divide both sides of equation by 1000:
2 = e^(.07*x)
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logarithms can be used to solve.
basic rule:
y = a^x if and only if log.a(y) = x
where log.a means log to the base a.
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therefore:
y = e^x if and only if log.e(y) = x
log.e = ln
equation becomes:
y = e^x if and only if ln(y) = x
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substituting 2 for y and .07*x for x, we get:
2 = e^(.07*x) if and only if ln(2) = (.07*x)
from the calculator, ln(2) = .693147181
equation becomes:
.693147181 = .07*x
x = .693147181/.07 = 9.902102579
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number of years for money to double with continuous compounding is 9.902102579 years.
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to prove, plug that value in the original equation.
2 = e^(.07*x)
becomes:
2 = e^(.07*9.902102579) = e^(.693147181) = 2
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formula checks out ok.
you can use your calculator to prove.
you can also use online continuous compounding calculator found at:
http://www.moneychimp.com/articles/finworks/continuous_compounding.htm
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