1. Finding x

We were given a couple of bonus questions the other day, and I'm not exactly sure how to get them.

The first one:

Find the value of 2^2^2... meaning it is unlimited and goes on forever. I know you would set it equal to x, but not sure afterwards.

The second one:

2 = x^x^x... it is unlimited and goes on forever.

Any help is appreciated.

2. Originally Posted by Shapeshift
We were given a couple of bonus questions the other day, and I'm not exactly sure how to get them.

The first one:

Find the value of 2^2^2... meaning it is unlimited and goes on forever. I know you would set it equal to x, but not sure afterwards.
if it goes on forever, then it will have no value. it tends to infinity. if you try solving it, you will get a solution that makes no sense.

The second one:

2 = x^x^x... it is unlimited and goes on forever.

Any help is appreciated.
Hint: log both sides

3. Originally Posted by Shapeshift
2 = x^x^x... it is unlimited and goes on forever.
This is the one that you can solve:

. . . . .$\displaystyle 2\, =\, x^{x^{x^{...}}}$

Following the hint provided previously:

. . . . .$\displaystyle \ln{(2)}\, =\, x^{x^{x^{...}}}\ln{(x)}$

But the first factor on the right-hand side is known to be equal to 2, so:

. . . . .$\displaystyle \ln{(2)}\, =\, 2\ln{(x)}$

. . . . .$\displaystyle \frac{1}{2}\ln{(2)}\, =\, \ln{(x)}$

. . . . .$\displaystyle \ln{(\sqrt{2})}\, =\, \ln{(x)}$

I'll bet you can take it from there! :wink:

4. .......

Hi,

Is it true ? : my answer :
X^X^X^X^X^X^...=X^(X^X^X^X^X^....) (1)
(X^X^X^X^X^X^...)=2 (2)
________________________________________
(1) & (2) => x^2=2 => x=2^(1/2)

bye

5. Originally Posted by c_o_d
Hi,

Is it true ? : my answer :
X^X^X^X^X^X^...=X^(X^X^X^X^X^....) (1)
(X^X^X^X^X^X^...)=2 (2)
________________________________________
(1) & (2) => x^2=2 => x=2^(1/2)

bye
when you say (1) & (2), what did you do? add them, subtract them...?

6. ......

I Use 2 instead of (X^X^X^X^...) ...