How the heck would you solve for n in the compund interest

Let's just assume n is a positive number≠1 because it's wierd otherwise

>>8524694
By filling in the other variables and doing linear algebra.

>>8524694
Fuck you're right, this kinda sucks to do out. Probably some crazy algebra shot that I'm not confident enough to work out.

Divide A by P.
Take log. Log base (1+r/n) (a/p)=n
Sorry I can't upload a picture from my phone. it's way better to see it.

>>8526144
>Log base (1+r/n) (a/p)=n
http://www.wolframalpha.com/input/?i=Log+base+(1%2Br%2Fn)+(a%2Fp)%3Dn

?

>>8524694
Take the ln of both sides. You can now move the exponent nt down and divide to solve for n. You will need number isk values for all other variables to get an answer, tho. Brainlet

>>8526152
I would use the change of base formula. Using a term with two variables as a base just feels like something i'd have to confess to a priest.

http://www.financeformulas.net/Solve-for-Number-of-Periods-PV-and-FV.html

>>8524694
In a real world situation n will always be known so don't worry about it

>>8526207
It's the only way I now how.
The change of base formula is too much to memorize, IMO. You have to remember that you'll know the values of P,A, and N. I can't see the importance of just algebraicly finding for n. If you do have a problem like that, it's maybe what? Once. Ever? lol

>>8526581
see>>8526284

somebody is in a Maths B course

>>8526460
Paying off debt in the shortest amount of time possible is a real world situation, faggot.

>>8524694
To solve for n, you need the product log function. Which is a bit of specific maths to be honest. I kinda doubt you'll need to solve what you've got there. That said, the output is:
http://www.wolframalpha.com/input/?i=a%3D(1%2Bb%2Fx)%5Ex+solve+x

[math]n = \frac{r}{W(\frac{e^{r+1}P}{A})-1} [/math]

Where W is the Lambert W """"function"""". I am not that confident in my algebra, though. But the steps are simple, make a substitution for [math]u = 1+\frac{r}{n} [/math] and try to make it look like [math] u e^{u}=f [/math] , then you lambertW both sides and substitute back the u.

A more practical solution that can be executed on even the lousiest calculators:

Let y = (A/P)^(1/rt), x = n/r (which we can pragmatically assume to be > 0), and the equation rearranges to y = (1 + 1/x)^x.

This is a strictly increasing function and so you can test various values of x until you narrow it down to a sufficient accurate interval, e.g. n/r <= x < [n+1]/r if you need it down to the nearest whole number.

In particular,

y = 2 --> x = 1
As x -> 0, y -> 1
As x -> + infinity, y -> e