# Generalise $y = pb^k$ where $k = c(1-x^a)$ given values for $p$, $b^c$ and $a$

I'm working on a pricing algorithm for work and I've reached a point where the rust on my brain is blocking the creative juices. I did a maths degree years ago, obviously deserving the odd fails I recorded, as I can't remember the rules for expanding exponents to determine whether this equation is solvable.

Suppose we have the following equation:

$y = pb^{c(1-x^a)}$

Both $p$ and $a$ are input values (base price and discount aggression respectively).

$b^c$ is a constant derived from two other input values (not listed - maximum discount, and minimum units for maximum discount)

So, we can assume that $p$, $a$, and $b^c$ are all constant values.

Can we rewrite this equation so that it can be generalised to $y=f(x)$ form?

I solved this to be $y = pe^{(1-x^a)\ln(b^c)}$, but it turns out an earlier equation that fed into this one was wrong. Back to the drawing board for me :)
• If you feel that rewriting $pb^{c(1-x^a)}$ as $pe^{(1-x^a)\ln(b^c)}$ answers the Question, then perhaps you should edit the Question and Answer to explain to Readers how this amounts to a "generalized... form". Both seem to give expressions $y = f(x)$, and since they are related by equality, the notion of generalizing seems unattained. – hardmath Mar 16 '17 at 1:01