The Christmas equation

I just posted the following on this year’s FPM Piazza forum.

Hi everyone,

I saw this a few years ago on QI, and found another version on the web at

If we are told that

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

multiplying by $r^2$ gives

$yr^2=\ln(\frac{x}m-sa)\,.$

Then taking $e$ to the power of both sides gives

$e^{yr^2}=\frac{x}m-sa\,,$

and multiplying both sides by $m$ gives

$me^{yr^2}=x-msa\,,$

which can be rewritten as

$me^{rry}=x-mas\,.$

(And a Happy New Year!)

Best wishes,

Dr Feinstein

Explaining mathematics