 Ask Remember Me? DrJ Posts: 1,328, Reputation: 339 Ultra Member #1 Aug 22, 2006, 02:22 PM
rearrange this formula to solve for n (no, its not my homework lol)
p = x*(1 - (1+r)^-n)/r

In words: take the negative nth power of (1+r). Subtract from 1,
divide by r. This gives the ratio of p to x.

I need to be able to solve for n.

Can someone help me with creating a formula out of this to solve for n? :confused: DrJ Posts: 1,328, Reputation: 339 Ultra Member #2 Aug 22, 2006, 04:08 PM
C'mon... I know there are some mathematical genius's out there... and I don't want to fall off the new posts page :cool: worthbeads Posts: 538, Reputation: 45 Senior Member #3 Aug 22, 2006, 05:16 PM
this one's a tough one. I'm not exactly sure about the answer, but if it helps, I think p=-x, but I would check that if I were you. DrJ Posts: 1,328, Reputation: 339 Ultra Member #4 Aug 22, 2006, 06:02 PM
Well, here's what I came up with:

-(1+r)^(1-rp/x)=n dmatos Posts: 204, Reputation: 26 Full Member #5 Aug 22, 2006, 06:27 PM
You have to use logarithms to "undo" powers. This is what I get:

n = ln(1-rp/x)/ln(1+r)

Where ln is the natural logarithm. A logarithm in any base would work, as long as you used the same top and bottom, but because of the special features of log(base e), ln is often used for this type of thing. rudi_in Posts: 251, Reputation: 45 Full Member #6 Aug 22, 2006, 06:47 PM
I agree that you will have to use logs to solve this one.

A quick verification on the calculator to show that this does work could be as follows...

2ⁿ = 8

n = (log 8) ÷ (log 2)

or

n = (ln 8) ÷ (ln 2)

n = 3 DaveS002 Posts: 2, Reputation: 1 New Member #7 Oct 28, 2006, 06:32 AM
Mathcad gives the following solution: -ln(-(p*r-x)/x)/ln(1+r).

Of course p, r, and x will have to be in proper bounds in order to take the ln.

Dave

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