Question: How can I algebraically solve $n-\log(na)=b$ for $n$. I figured out a simple algorithm to get arbitrary close to the solution (estimate, then negate log term and use answer in the log term, then repeat), but I want to know how I can solve this exact.

Answer: 
You can't solve this algebraically. However, you can rewrite your equation
and express it in terms of a known special function, the
[Lambert W function](http://en.wikipedia.org/wiki/Lambert_W_function):

$$n - \log(na) = b \iff n - \log n = b + \log a \iff \frac{e^n}{n} = a e^b
\\ \iff n e^{-n} = \frac{1}{a e^b} \iff (-n)e^{-n} = -\frac{1}{a e^b}
 \implies n = -W\left(-\frac{1}{a e^b}\right)$$
**Hint:** Use the [Lambert W-function](http://en.wikipedia.org/wiki/Lambert_W_function).