Question

Any clues on how to best solve for x in terms of a,b, and c:
((x-1/2)/(x(1-a)/a-1/2))^(-a)=(e^b ((x^2 (1-a))/a^2 -x(2-a)/2a+1/4))/e^c(1-a)

Answer 1

\[\huge(\large \frac{x-\frac{1}{2}}{\frac{x(1-a)}{a}-\frac{1}{2}}\huge)\large^{^{-a}}\large =\frac{e^b (\frac{x^{2} (1-a)}{a^{2}} -\frac{x(2-a)}{2a}+\frac{1}{4})}{e^{c(1-a)}}\]

Answer 2

\[\large \frac{x-\frac{1}{2}}{\frac{x(1-a)}{a}-\frac{1}{2}}\large =\frac{e^{ab}}{e^{c(1-a)a}} (\frac{x^{2} (1-a)}{a^{2}} -\frac{x(2-a)}{2a}+\frac{1}{4})^{a}\]

Answer 3

It is from trying to simplify the process of estimating parameters from a compound gamma distribution. Just thought I might get lucky and find someone here who has mad algebraic skillz.

Answer 4

Bummer. Thanks for trying. Oh, unrelated, but fun fact I just learned today:
\[\huge e^{\frac{\frac{d (x!)}{dx}}{x!}}=x+\frac{1}{2}\]

Answer 5

@pgpilot326 @ganeshie8 @wio

Answer 6

\[\huge e^{\left(\frac{\frac{d (x!)}{dx}}{x!}\right)}\large=x+\frac{1}{2}\]