SOLUTION: How long will it take an investment to double if it is invested at a 5% annual interest rate and compounded continuously?
So, I'm really drawing a complete blank on this, i know
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Question 313164: How long will it take an investment to double if it is invested at a 5% annual interest rate and compounded continuously?
So, I'm really drawing a complete blank on this, i know P=e^rt, so P=e^0.05t, but that's far as I got. Thanks!
Found 2 solutions by solver91311, nerdybill:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
So far, so good, except for one tiny detail. The formula you want is:
Where 
is the amount at the end, 
is what you start with, 
is the rate as decimal, 
is the time in years, and 
is the base of the natural logarithms.
Since you want your money to double,
Therefore:
You don't need to substitute the rate value just yet...patience, Grasshopper.
Take the natural logarithm of both sides:
\ =\ \ln(2))
Next apply the following rule:
\ =\ n\log_b(x))
\ =\ \ln(2))
Next use the fact that \ =\ 1)
)
Since you want to know , multiply both sides by 
}{r})
NOW substitute the value of
}{0.05})
And finally get all happy with your calculator
years.
By the way, this same formula will work for "How long will it take my money to multiply by 
at interest rate compounded continuously.
}{r})
John
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
How long will it take an investment to double if it is invested at a 5% annual interest rate and compounded continuously?
.
You forgot the "initial principal" in your formula...
.
Continuously compounded interest:
P=(Po)e^rt
Where
P is the amount after time t
Po is the initial investment
r is the rate
t is the time
.
So, the "trick" is to assign a variable to the intial investment:
Let x=initial investment (Po)
So, when we double it we should have:
2x = final amount (P)
.
Plug in what we know into:
P=(Po)e^rt
2x = xe^(.05t)
Now, we solve for t:
We begin by dividing both sides by x:
2 = e^(.05t) (notice our variable is gone!)
ln(2) = .05t
ln(2)/.05 = t
13.86 years = t
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