Question 303556: An investment with interest compounded continuously doubled itself in 16 yr. What is the interest rate? Please help, I don't get this word problem at all.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Continuous compounding formula is:
F = P * e^(i*y)
F = Future Value
P = Present Value
e = the scientific constant of 2.718281828
i = the annual interest rate.
y = the number of years.
You want to know what the interest rate is if the money doubles every 16 years.
In the equation above:
F = 2
P = 1
e = the scientific constant of 2.718281828
i = i
y = 16
Since the money doubles, then the future value has to be twice the present value which is why we chose F = 2 when P = 1.
The formula becomes:
2 = 1 * e^(16*i)
This is the same as:
2 = e^(16*i)
You want to find i.
Take the log of both sides of this equation to get:
log(2) = log(e^(16*i))
Since, in general, log(x^y) = y*log(x), then your equation becomes:
log(2) = 16*i*log(e))
Divide both sides of this equation by (16*log(e)) to get:
i = log(2) / (16*log(e))
Use your calculator to find log(2) and log(e).
the equation becomes:
i = .301029996 / (16 * .434294482) which becomes:
i = .301029996 / 6.94871171 which becomes:
i = .043321699
That's your annual interest rate.
Plug that back into your original equation to confirm that it is good.
Your original equation is:
2 = 1 * e^(16*i)
replace i with .043321699 to get:
2 = 1 * e^(16*.043321699) which becomes:
2 = 1 * e^(.693147181) which becomes:
2 = 1 * 2 which becomes:
2 = 2 which is true, confirming that the answer is good.
Your answer is that the annual interest rate is equal to .043321699 which is equivalent to 4.33% per year rounded to the nearest hundredth of a percent.
To solve this problem, you had to know the continuous compounding formula which I gave you above.
You also had to know that you needed to take the log of both sides of the equation in order to solve it.
You also had to know that log(x^y) = y*log(x).
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