Thread: Amortization - LOGS!! Help

I'm actually working on an amortization problem.

I am trying to solve for n. I have all of the other values.

In order to do this I end up using logs (i think...)

A/P/r = (1+r) ^n /((1+r)^n - 1)

Does that turn into something like...

log[A/P/r] = log[(1+r)^n] - log[(1+r)^n - 1]

And where do I go from there?!?

THANKS!!

You want to solve for n from the
A = P[r*(1+r)^n] / [(1+r)^n -1].

Here is one way. The idea is just to isolate n by itself alone on one side of the equation.

Divide both sides by r*P,
A / rP = [(1+r)^n] / [(1+r)^n -1]

Multiply both sides by [(1+r)^n -1],
[A / rP]*[(1+r)^n -1] = (1 +r)^n

Expand,
[A / rP](1+r)^n -[A / Pr] = (1+r)^n

Collect like terms,
[A / Pr](1+r)^n -(1+r)^n = [A / Pr]

(1+r)^n *{[A / Pr] -1} = [A / Pr]
(1+r)^n = [A / Pr] / {[A / Pr] -1}

Take the natural logs of both sides,
n*Ln(1+r) = Ln[A / pr] -Ln{[A / Pr] -1}

Therefore,
n = (Ln[A / Pr] -ln{[A / Pr] -1}) / Ln(1+r)