Compound interest problem

thebristolsound · #1 $Old$ November 6th, 2009, 09:00 PM

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

Prove It · #2 $Old$ November 6th, 2009, 09:34 PM

Quote:

Originally Posted by thebristolsound $View Post$

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

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

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

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

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

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

$n = \frac{\ln{2}}{\ln{1057} - \ln{1000}}$ .

thebristolsound · #3 $Old$ November 6th, 2009, 10:31 PM

I got the answer 12.5038 and it wasn't quite correct, did I miss a step somewhere?

Prove It · #4 $Old$ November 6th, 2009, 11:04 PM

It asks you for a whole number of years.

So if it's a little more than 12...

thebristolsound · #5 $Old$ November 6th, 2009, 11:13 PM

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?

Prove It · #6 $Old$ November 6th, 2009, 11:21 PM

Can't you round this to 3 decimal places?

$12.5038 \approx 12.504$ .

thebristolsound · #7 $Old$ November 6th, 2009, 11:28 PM

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

Prove It · #8 $Old$ November 6th, 2009, 11:31 PM

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...

thebristolsound · #9 $Old$ November 6th, 2009, 11:35 PM

I'm going to try and use the formula A = Pe^rt I think this is what it's looking for

Guess992 · #10 $Old$ November 7th, 2009, 05:08 AM

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.

Prove It · #11 $Old$ November 7th, 2009, 05:41 AM

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

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

$= Pe^{rt}$ .

thebristolsound · #12 $Old$ November 7th, 2009, 10:46 AM

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

if money is invested at

percent and compounded continuously, then the factor

multiplying the initial investment after

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

yields

which with our value of

percent gives a value of