# Need help finding the doubling time

• Mar 17th 2010, 01:29 PM
florx
Need help finding the doubling time
Suppose you invest $1500 in an account that pays interest at a nominal rate of 4.5% per year compounded monthly. a) How long will it take the investment to double? b) What continuous annual interest rate would result in the same annual yield? Please help with these these problems. For a) I have set up the problem as 3000 = 1500(x)^t, solve for t, however I am stuck as to whether the x would be the effective rate or the nominal rate? Would I put in 1.045 (nominal rate) for x and solve for t or would I put in 1.045939825 (effective rate) in for x and solve for t? For b) I don't even understand what they are asking for. Are we trying to find what is the rate and how would we set up this problem? On the left side would it be the formula 1500e^r*t = (something)? Please shed some light on this problem also. Thank you so much • Mar 17th 2010, 01:41 PM e^(i*pi) Quote: Originally Posted by florx Suppose you invest$1500 in an account that pays interest at a nominal rate of 4.5% per year compounded monthly.

a) How long will it take the investment to double?
b) What continuous annual interest rate would result in the same annual yield?

For a) I have set up the problem as 3000 = 1500(x)^t, solve for t, however I am stuck as to whether the x would be the effective rate or the nominal rate? Would I put in 1.045 (nominal rate) for x and solve for t or would I put in 1.045939825 (effective rate) in for x and solve for t?

For b) I don't even understand what they are asking for. Are we trying to find what is the rate and how would we set up this problem? On the left side would it be the formula 1500e^r*t = (something)? Please shed some light on this problem also.

Thank you so much

The compound interest formula is (shamefully copy/pasted from wikipedia)

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

Where,

* P = principal amount (initial investment)
* r = annual nominal interest rate (as a decimal)
* n = number of times the interest is compounded per year
* t = number of years
* A = amount after time t

P = 1500
r = 0.045
n = 12 (there are 12 months in a year)
A = 2P
t = t

Part A is asking for what value of t does $A = 2P$. Sub in your values and find t.

Continuously compounded interest means we can use $e$ in part B and the formula becomes $A = Pe^{nt}$. In this case you are solving for n.
• Mar 17th 2010, 03:39 PM
florx
Quote:

Originally Posted by
Continuously compounded interest means we can use [tex
e[/tex] in part B and the formula becomes $A = Pe^{nt}$. In this case you are solving for n.

So the equation would be 1500(1 + 0.045/12)^12 = 1500e^12*n? And also should't we be solving for r since the equation is asking What continuous annual interest rate would result in the same annual yield?
• Mar 17th 2010, 03:47 PM
e^(i*pi)
Quote:

Originally Posted by florx
So the equation would be 1500(1 + 0.045/12)^12 = 1500e^12*n? And also should't we be solving for r since the equation is asking What continuous annual interest rate would result in the same annual yield?

I made a mistake in my notation. For part b $n = r$

That equation looks fine to me
• Mar 17th 2010, 04:01 PM
florx
So we are solving for r correct?

New formula should be 1500e^r(1) = 1500(1 + 0.045/12)^12?