I have almost isolated $x$ in the equation:
$\displaystyle \frac{\frac{z}{\ln\left(2\right)}at^{a}z^{a}\left(z+1-t\right)}{\left(z+1-t\right)^{a+1}}=x^{a+1}\left(\left(\frac{tz}{x\left(z+1-t\right)}\right)^{a}-1\right)$
Which may be simplified to the general expanded form (where $a,b$ & $c$ are constants):
$c=x^{a+1}\left(\frac{1}{x}b\right)^{a}-x^{a+1}$
How would I go about solving for $x$ here? I have been stuck for a while and would grateful for any help.
$$c=x^{a+1} (\frac{1}{x} b)^a - x^{a+1}$$ $$c=x^{a+1} (\frac{1}{x^a} b^a) - x^{a+1}$$ $$c=x b^a - x^{a+1}$$ Let $x=by$ (the problem when $b=0$ can be done) $$c=(by)b^a - (by)^{a+1}$$ $$c=yb^{a+1} - b^{a+1} y^{a+1}$$ Let $d=a+1$ $$\frac{c}{b^d} = y - y^d.$$ $$y^d -y +f =0$$ Even if $d$ is an integer, there is no way in general to invert $d$th order polynomials, however there may be a technique for this specific polynomial. You can say that there is a solution if $d$ is odd ($a$ is even).