Compound interest problem

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November 6th 2009, 07:00 PM
thebristolsound

Compound interest problem

You are investing money at 5.7 percent annual interest, compounded continuously. It will take you how many years to double your investment?
November 6th 2009, 07:34 PM
Prove It

Quote:

Originally Posted by thebristolsound

You are investing money at 5.7 percent annual interest, compounded continuously. It will take you how many years to double your investment?

The formula for compount interest is

$A = P(1 + r)^n$ .

Here, $r = 5.7\%, A = 2P$

So $2P = P\left(1 + 5.7\%\right)^n$

$2 = \left(1 + \frac{5.7}{100}\right)^n$

$2 = \left(\frac{1057}{1000}\right)^n$

$\ln{2} = \ln{\left[\left(\frac{1057}{1000}\right)^n\right]}$

$\ln{2} = n\ln{\left(\frac{1057}{1000}\right)}$

$n = \frac{\ln{2}}{\ln{\left(\frac{1057}{1000}\right)}}$

$n = \frac{\ln{2}}{\ln{1057} - \ln{1000}}$ .
November 6th 2009, 08:31 PM
thebristolsound

I got the answer 12.5038 and it wasn't quite correct, did I miss a step somewhere?
November 6th 2009, 09:04 PM
Prove It

It asks you for a whole number of years.

So if it's a little more than 12...
November 6th 2009, 09:13 PM
thebristolsound

It actually asks me to be within one tenth of one percent so 3 numbers beyond the decimal point. I'm sorry for not specifying that. Did I just miss something in my calculations?
November 6th 2009, 09:21 PM
Prove It

Can't you round this to 3 decimal places?

$12.5038 \approx 12.504$ .
November 6th 2009, 09:28 PM
thebristolsound

Yeah, it's an online homework problem and the system isn't accepting it, for some reason

WeBWorK : math1050fall2009-1 : 9 : 1

I don't know if you can access that, but thank you so much for the help
November 6th 2009, 09:31 PM
Prove It

What does it mean by "compounded continuously?"

How often does it get compounded... I assumed that it was every year, since the interest rate is given per year, but the wording makes me think otherwise...
November 6th 2009, 09:35 PM
thebristolsound

I'm going to try and use the formula A = Pe^rt I think this is what it's looking for
November 7th 2009, 03:08 AM
Guess992

Continuous interest is calculated using $Pe^{rt}$

@Prove It
Continuous interest is basically compounded interest that is compounded over such small periods of time, that it is continuous.
November 7th 2009, 03:41 AM
Prove It

Yes, I just did some research on that.

It seems that when the number of times the interest is compounded $\to \infty$ , the interest tends to

$\lim_{t \to \infty}P(1 + r)^t$

$= Pe^{rt}$ .
November 7th 2009, 08:46 AM
thebristolsound

Thanks guys, after finding out that formula I was able to solve it like this

if money is invested at http://webwork1.math.utah.edu/webwor...da050b4dd1.png percent and compounded continuously, then the factor http://webwork1.math.utah.edu/webwor...ebc71269f1.png multiplying the initial investment after http://webwork1.math.utah.edu/webwor...5b59f2bc11.png years is given by
http://webwork1.math.utah.edu/webwor...1047150cf1.png
To find the time at which the initial investment is doubled we solve the equation
http://webwork1.math.utah.edu/webwor...9418750671.png
Taking the natural logarithm on both sides and dividing by http://webwork1.math.utah.edu/webwor...79cb7891c1.png yields
http://webwork1.math.utah.edu/webwor...28beb230d1.png
which with our value of http://webwork1.math.utah.edu/webwor...bfa28ecf81.png percent gives a value of
http://webwork1.math.utah.edu/webwor...5d285c8e21.png
December 11th 2009, 02:41 PM
Pannoowau

Still didn't invested but in plane to invest.....

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Compound Interest Formula