$$\Large x^{x^{x^{x^{x^{.^{\,.^{\,.}}}}}}} = 2$$  What is $x$?
122151
How do you approach a problem like (solve for $x$):  $$x^{x^{x^{x^{...}}}}=2$$  Also, I have no idea what to tag this as.   Thanks for any help.
28282
Assuming that $e&lt;y&lt;x$, prove that $ x^y &lt; y^x$.  I think this must be easy, but I can't work it out. Thanks in advance for any kind of help.
13169
If values $a$, $b$, and $c$ are known, is there an efficient way to find $x$ in the equation: $a^{x} \bmod b = c$?  E.g. finding $x$ in $128^{x}\bmod 209 = 39$.
107414
I know  that uses $x={\sqrt{2}}^{\sqrt{2}}$ to prove that there must exist an irrational to an irrational power that evaluates to a rational. But I don't know if $x$ itself is known to be irrational or rational.  Could someone enlighten me? Thanks!
13240
Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases.
13294
Say you have a number $x^{\sqrt 2}$.  Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this.  I was thinking something like $$x^{2^{1/2}}$$ and extending on that idea, but I don't know how.  Also, what about for $x^{\sqrt 3}$? Or $x^{\sqrt n}$?
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Can someone please explain me what is a ?  Also, how does putting $$X_a(\psi) ~=~ x_a\star\psi$$ realise $$[X_a,X_b]=i\theta_{ab}{\bf 1}?$$   Ref:
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Out of boredom, I decided to recall the following equation:   $$x^{x^{x\cdots}} = 2.$$  Which, I simply rewrote like this:   $x^2 = 2$, and therefore $x = \sqrt{2}$.  Then I took a look at the more general form:  $$x^{x^{x\cdots}} = p.$$  I then concluded that $x = (p)^{\frac{1}{p}}$, and this was the solution set for all $p$. Then I thought for a while and determined that there are multiple values where this function equals the square root of $2$, so its not injective. However, this led to an awkward statemend based on the fact that $\sqrt{2} = (4)^{\frac{1}{4}}$. We deduce the following:  The equation $x^{x^{x\cdots}} = 2$ and the equation $x^{x^{x\cdots}} = 4$ have the same solution, namely $x = \sqrt{2}$. So we conclude that $2 = 4$.   I showed this to friend of mine, and he conjectured that there are no valid solutions for all $p &gt; e$. To prove this, I tried writing the following:   I tried differentiating $f(x) = (x)^{(1/x)}$ and setting it equal to $0$ to find the maximum. I determined that there was a maximum at the point where $x = e$. Does this mean that the maximum possible solution $x$ for the infinite power is $e^{(1/e)}$ therefore showing that the maximum value for $p$ is indeed $p = e$?
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Possible Duplicate:          I obtained a question asking for how to solve $\large x^y = y^x$. The given restraints was that $x$ and $y$ were both positive integers. By a bit of error an trial we quickly see that $x=2$ and $y=4$ is one solution.   My question is: How do one show that $(2\,,\,4)$ is the only non trivial, positive solution to the equation?   Now my initial approach was as follows: We have   $$\large x^y = y^x$$  The trivial solution is obviously when $y=x$, so let us focus on when $y \neq x$.  Let us make a more general statement. Firstly I take the log of both sides  $$\large y \log x = x \log y $$  Let us divide by x and \log x (We now assume $x\neq 0$ and $x\neq 1$ since 0 is not a positive number, and 1 gives us a trivial solution)  $$\large \frac{y}{x} = \frac{\log y }{\log x}$$  For these sides to be equal, we must remove the logarithms on the right hand side, this is achived if $y$ is on the form $x^a$. Now This gives  $$\large \frac{x^a}{x} = \frac{\log \left(x^a\right) }{\log(x)}$$  $$\large x^{a-1} = a$$  So finaly we obtain that $ \displaystyle \large x=\sqrt[ a-1]{a}$ and $\displaystyle \large y = \sqrt[a-1]{a^a}$  Now setting $a=2$ gives us $x = 2$ and $y=4$ as desired.  My question is, how do we prove that $x=2$ and $y=4$ is the only integer solutions? My thought was to show that $ \displaystyle \large \sqrt[ a-1]{a}$ and $\displaystyle \large \sqrt[a-1]{a^a}$ are both irrational when a>2, but I have not been able to show this.   Any help is greatly appreciated, cheers =)
17989
We can solve the equation $x^x=c$ using the Lambert $W$ function, but how do we solve $x^{x^x}=c$ ,
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Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x $?   I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.  Thank you.
12359
Is $2^{2n} = O(2^n)$?  My solution is:  $2^n 2^n \leq C_{1}2^n$  $2^n \leq C_{1}$,  TRUE.  Is this correct?
13008
Can someone help me solve the equation $x^2 - 1 = e^x$ ?  I tried taking the natural logarithm of both sides but I don't know where to go from there..  I got:  $\ln(x^2 -1) = x$ But I don't know how to solve it from here. Any help please?
33263
Could you help me with this one? Thanks. The answer should be 1, somehow. I tried everything I know, but I couldn't solve it.  $$\lim_{x\to0^+} (x^{x^x-1})$$
49558
If $x^{x^4}=4$; Then what is the value of $x^{x^2}+x^{x^6}$ ;  I just want a hint like from where should I start. (other than using a trial and error to find that $x=\sqrt{2}$ satisfies this and now just put it in the required equation!!)
32818
Using the  for $\Gamma(x)$ and $\Gamma\Big(x + \frac12\Big)$, how can I prove the ? This is problem $10.7.3$ in the book , by Boros and Moll.  Any help is highly appreciated.
39492
What is the value of $1^i$? $\,$