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I stumbled across the equation,

$a^x=\Gamma(x) \quad \mathrm{for} \quad a \geq 1$

while trying to prove that $a^{n!}$ eventually becomes larger than $(a^n)!$ for sufficiently large $n$. Specifically for $n$ such that $a^n < (n-1)!, \quad a^{n!}>(a^n)!$. While that particular $n$ isn't tight (i.e., the inequality reverses before that value of $n$), the solution to the above equation is sufficiently large to guarantee the inequality.

The function $f(x,a)=a^x-\Gamma(x)$, with $a \geq 1$, has exactly two roots, and I'm interested in the one larger than $1$. I'm having difficulty in finding the solution numerically since the function cuts the x-axis in an almost perpendicular fashion, making the root-finding heavily dependent on the initial value (and slope of the intersection becomes steeper with increasing $a$). For instance, here is $f(x,4)$:

plot of f(x,4)

The roots of $f(x,4)$ are (approximately): $x=0.46488, \, 11.1489$

Also, could the functional equation, $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin{\pi x}}$, help? Or maybe some sort of inverse-gamma function?

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    I would like to amplify on the important comment by J.M. The ever popular Newton Method is often practically much less efficient than the Secant Method. True, Secant Method usually needs more iterations, but the iterations are often much cheaper, particularly if the derivative has to be evaluated numericallly.2011-04-18

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