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

+ 0 like - 0 dislike
83 views

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.)
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.