Die Lösung der Weihnachtsgleichung

$~~~~~~y~=~{\ln(x/m ~-~ sa)}/r^2$

$\doubleleftright~ r^2y~=~\ln(x/m) ~-~ as$

$\doubleleftright~ \e^{r^2y} ~=~ x/m ~-~ as$

$\doubleleftright~ \e^{r^2y} ~+~ as ~=~ x/m$

$\doubleleftright~ m(\e^{r^2y} ~+~ as) ~=~ x$

$\doubleleftright~ m\e^{r^2y} ~+~ ams ~=~ x$

$\doubleleftright~ m \e^{r^2y} ~=~ x~-~ams$

also

$m \e^{rry} ~=~ x-mas$