# Is this equation hard to solve?

How do I solve the equation below for $x$? $A$, $B$, $a$ and $r$ are constants.

$$x + \frac{1}{1+Ar^{-x/a}} + B =0$$

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I assume you meant to post this on the main site, not here on meta. –  Ilmari Karonen Mar 4 at 10:24
Instead of closing, I vote for migration to our main site. –  Jyrki Lahtonen Mar 4 at 10:33
@JyrkiLahtonen: I would've done that, but couldn't find the option. Did I just miss it, or can we no longer migrate posts to the main site from meta without bothering the mods? –  Ilmari Karonen Mar 4 at 10:36
How could i transfer the question to main site? –  Norman Mar 4 at 10:45
I don't think you can solve it analytically. But sure you can numerically, for given values of $a,A,B$ and $r$ of course. –  Kaster Mar 4 at 11:26
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## migrated from meta.math.stackexchange.comMar 4 at 10:59

This question came from our discussion, support, and feature requests site for people studying math at any level and professionals in related fields.

Even in the simple case $$x+\frac{1}{1+e^{-x}}=0$$ with solution $x \approx -0.401058137541547$, it is identified by the ISC only as "solution of $z+W(z+1)=0$" where $W$ is the Lambert W function.
Once you supply the particular values of $A , B , a$ and $r$ you can use a black box to solve it like Wolfram alpha, or you can use a numerical method like Newton's method to solve it.