# X^x^x^x^x = 2

1. Aug 18, 2010

### starfish99

In a book of math puzzles Peter Winkler discusses the sequence x, x^x, x^x^x, x^x^x^x and writes about conditions for its divergence. Clearly for x=1 the sequence is 1,and for x=2 it diverges to infinity.

Then he shows that the maximum value of x for the sequence to converge is x=e^(1/e) or
x= 1.44467... At this value the infinite tower of "x" exponents is equal to e (2.7182818..).
For any x larger than e^(1/e), the sequence diverges to infinity.

Mr Winkler later goes on to discuss the equation x^x^x^x^x...=2 (an infinite tower of "x"
exponents=2)
By using the trick that the exponent of the bottom"x" is the same as the whole expression,the equation becomes x^2=2, and x=sqrt(2)=1.414... is the solution.
(This is close to the maximum value for convegence( shown above) 1.44467...

My question(finally):
Suppose you have an equation x^x^x^x^x^x^x...=10 (an infinite tower of "x"=10)
Why can't you use the same trick as we did for x^x^x^x^x^x=2 case.
In this case we would get x^10=10. And the solution is x= the tenth root of 10(x=1.2589..)
Now the tenth root of 10 is clearly the wrong answer because:
1) It is too small . It is smaller than sqrt(2) whose tower conveges to the number 2
2) The maximum value this tower of "x" converges to is 2.71828... at x=e^(1/e)

Why doesn't this trick work for x^x^x^x^x^x...=10 ?

2. Aug 18, 2010

### hamster143

Re: x^x^x^x^x...=2

Because there is no x for which the sequence converges to 10? The trick works for small x because we presume that the solution exists, which is not the case here.

3. Aug 18, 2010

### starfish99

Re: x^x^x^x^x...=2

Thanks hamster. Sometimes you can forget, some equations have no solutions.