SAP forum上的万能解，所有数都能生成

Regarding the IBM monthly puzzle, my solution is
LN(√(EXP(LN(2)/LOG(2))))

a friend of mine just sent me another answer
sec(arctan(2))^2 = 5

it is good. However, not many people have heard of sec(). We may revise it as:
exp(-ln(cos(arctan(2))^2))

And we can have another approach:
first: √√√√√√exp(2)
then ln (√√√√√√exp(2))=ln(exp(2))/64
ln (ln(√√√√√√exp(2)))= ln(ln(exp(2)))-ln64=ln2-6ln2=-5ln2
-ln (ln(√√√√√√exp(2)))/ln2=5

所有跟帖： 

• 你，另一个天才！ -戏雨飞鹰- ♀ (0 bytes) () 01/12/2010 postreply 12:02:39

• 回复：SAP forum上的万能解，所有数都能生成 -Jinjing- ♀ (103 bytes) () 01/12/2010 postreply 21:00:59