OldChinaHand  
Oct 05, 2006
| Like homework from Hades...  |
zigthepig 
Oct 05, 2006
| Agh, you know I can't do these! I don't even understand googol!  Nice one.  |
(user deleted)
Oct 07, 2006
| 2 and 4. By the way,how do I earn points? |
(user deleted)
Oct 07, 2006
| I need 250 points!  Help! |
pating
Oct 09, 2006
| quite easy, but fun nontheless.
the problem with these kind of teasers is that it has no real solution. it trial-and-error. so if you got the answers right, its always fun. but if you dont, its frustrating.
sometimes you just have to rely on luck and instinct.
its more challenging this way! |
keveffect1
Oct 29, 2006
| adeline, you earn one point for each time you submit a vote on a teaser. So vote on 250 of them and you will be sitting pretty |
(user deleted)
Oct 30, 2006
| have you proof that 2 and 4 are the only two distinct reversible exponents? what about -2 & -4? |
qwertyopiusa
Nov 23, 2006
| Trial and error?
I disagree.
In few steps you can rewrite the equation as Ln[r]/r=t.
The function Ln[r]/r is only positive for r>1, and it has the maximum 1/e at r=e.
So, for each value of t with 0 |
qwertyopiusa
Nov 23, 2006
| So, for each value of t with 0 |
qwertyopiusa
Nov 23, 2006
| hmmm... it doesn't want to paste the whole answer... < is this character the problem? |
qwertyopiusa
Nov 23, 2006
| So, for each value of t between 0 and 1/e, the above equation has 2 solutions, x and y, each of them in the intervals [1,e] and [e,Infinity) respectively. In other words, there are infinite solutions for this problem. |
qwertyopiusa
Nov 23, 2006
| Assuming that the problem is restricted to the algebra of natural numbers, the only possible value for x between 1 and e is x=2 (e=2.718...). IF there is a solution at all, that solution is unique and it is with x=2. Replacing x=2 in the original equation gives y=4, and therefore the solution exist in the group of integer numbers.
Maybe the most challenging step is the last one, where you have to calculate one solution (let say y) as a function of the other one (x):
y(x)=ProdLog[-Ln[x]/x]/(-Ln[x]/x)
where w=ProdLog[z] is the well-known "logarithmic product" function (w is the solution of the equation z=w*e^w). |
javaguru
Jan 14, 2009
| I got two solutions:
2^4 = 4^2 = 16
(-2)^(-4) = (-4)^(-2) = 0.0625
|
