Solving for n in the exponent. Well, it's another question I feel like I should know. I'm trying to model the number of successes before the first failure. The probability of successes is given as $p$, which makes the probability of failure $(1-p)$.
The probability mass function, as I've calculated it, turns out to be $p^{n-1}(1-p)$, since we will stop at the first failure.
I'm trying to solve the following equation for n, but I'm at a loss for how to get it out of the exponent.
$1=p^{n-1}(1-p)$
I would appreciate any help anyone can give me.
Thanks

Using the equation you gave, $1 = p^n (1-p)$
You should get the term with "n" by itself:
$1/(1-p) = p^n$
and then simply take the logarithm
$-log(1-p) = n*log(p)$
$n = -log(1-p)/log(p)$
However, I think that the problem itself might be more complicated than you've realized. It is the sum of all possible values of "n" - not just one of them - from 0 to ∞ that would make the probability equal to 1.
$1 ≠ (p^n)(1-p)$
$1 = (1-p)$$(p^0+p^1+p^2+p^3+p^4+...)$
