# $me^{rry} = x -mas$ !

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Hi all, I wish ...

$\begin{eqnarray} y & = & \frac{\ln\left( \frac{x}{m} - sa \right)}{r^2} \\ yr^2 & = & \ln\left( \frac{x}{m} - sa\right) \\ e^{yr^2} & = & \frac{x}{m} - sa \\ me^{yr^2} & = & x - msa \\ me^{rry} & = & x - mas \\ \end{eqnarray}$

... to the whole PhysicsOverflow community ;-) !

(Also, am I the only one who got confused if the equation had any physical significance? Yeah, I know, taking the $r^2$th root within a logarithm is probably not going to be useful.)
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