How would I convert from frequency back to percent? I'm working on a web application that uses an oscillator. It has a parameter that goes from 0 to 100 percent. I translate that percent into hertz using this equation: 
$$
\text{frequency} = 2^{(\lfloor  128*\text{percent} \rfloor - 69)/12}*440
$$

I am having some trouble solving the equation for percent so that given a frequency, I can get a percent value to send back to my app. Help please? :]
$$f = 2^{(x - 69)/12} \times 440$$
inverts into
$$x = 69 + 12 \log_2(f/440)$$
but in your case $x = \lfloor  128p \rfloor$, which is impossible to invert exactly because it maps intervals to single numbers (i.e. the function is not one-to-one and so does not have an inverse).

For example, $p = 10^{-9}$ and $p = 10^{-8}$ would yield the same $x=0$, so if I tell you $x=0$, there is no way to imply what the original $p$ was directly.

EDIT
You could bound it, however. $x$ must certainly be an integer and then, since you are rounding down, $x = \lfloor  128p \rfloor$ means
$$
x \leq 128p  &lt; x+1 \Leftrightarrow \frac{x}{128} \leq p &lt; \frac{x+1}{128}
$$
Since the floor function is not bijective, one can not invert this equation. Ignoring the floor function gives:
$$p=\frac{69}{128}+\frac{12\ln\left(\frac{f}{440}\right)}{128\ln 2}$$
where $f$ is the frequency and $p$ is the percentage.