# Compound interest problem

• Nov 6th 2009, 08:00 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?
• Nov 6th 2009, 08:34 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 $2P = P\left(1 + 5.7\%\right)^n$

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

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

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

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

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

$n = \frac{\ln{2}}{\ln{1057} - \ln{1000}}$.
• Nov 6th 2009, 09:31 PM
thebristolsound
I got the answer 12.5038 and it wasn't quite correct, did I miss a step somewhere?
• Nov 6th 2009, 10:04 PM
Prove It
It asks you for a whole number of years.

So if it's a little more than 12...
• Nov 6th 2009, 10:13 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?
• Nov 6th 2009, 10:21 PM
Prove It
Can't you round this to 3 decimal places?

$12.5038 \approx 12.504$.
• Nov 6th 2009, 10:28 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
• Nov 6th 2009, 10:31 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...
• Nov 6th 2009, 10:35 PM
thebristolsound
I'm going to try and use the formula A = Pe^rt I think this is what it's looking for
• Nov 7th 2009, 04:08 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.
• Nov 7th 2009, 04:41 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

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

$= Pe^{rt}$.
• Nov 7th 2009, 09:46 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 To find the time at which the initial investment is doubled we solve the equation Taking the natural logarithm on both sides and dividing by http://webwork1.math.utah.edu/webwor...79cb7891c1.png yields which with our value of http://webwork1.math.utah.edu/webwor...bfa28ecf81.png percent gives a value of
• Dec 11th 2009, 03:41 PM
Pannoowau
Still didn't invested but in plane to invest.....

__________________
Compound Interest Formula