- #1

starfish99

- 28

- 0

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 ?