Which is bigger $x^{log(x)}$ OR $(log(x))^x$ I'm trying to find out which is bigger $x^{log(x)}$ OR $(log(x))^x$
As $x \to \infty$
I tried to take the log of both but I didnt't reach any where.
 A: Hint: $\dfrac{\ln y}{y}$ is decreasing for $y > e$.

 Hence $a^b < b^a$ for $e < b < a$.

A: You can try to take $\lim\limits_{x\rightarrow +\infty}x^{\log(x)}/\log(x)^x$. To do that, see that:
$$\begin{aligned}
\lim\limits_{x\rightarrow +\infty} \frac{x^{\log(x)}}{\log(x)^x} &=\lim\limits_{x\rightarrow +\infty}
\frac{\exp\left(\log(x)\log(x)\right)}{\exp\left(x\log(\log(x))\right)} \\
&= \lim\limits_{x\rightarrow +\infty} \exp\left(\log(x)\log(x)-x\log(\log(x))\right) \\
&= \exp\left(\lim\limits_{x\rightarrow +\infty}(\log(x))^2-x\log(\log(x))\right)
\end{aligned}$$
and so you have to compare $(\log(x))^2$ and $x\log(\log(x))$. So take
$$\lim\limits_{x\rightarrow +\infty}(\log(x))^2-x\log(\log(x))$$
This equals
$$\lim\limits_{x\rightarrow +\infty}x\log(\log(x))\left(\frac{(\log(x))^2}{x\log(\log(x))}-1\right)$$
which is a much easier limit to analyse. Applying l'Hôpital's rule to the fraction inside the parentheses we have 
$$\lim\limits_{x\rightarrow +\infty}\frac{2\log(x)/x}{\log(\log(x))+1/\log(x)}$$
As x approaches infinity, the numerator approaches 0 and the denominator approaches infinity, so the whole limit approaches 0. Therefore our original limit can be found because
$$\begin{aligned}
\lim\limits_{x\rightarrow +\infty} (\log(x))^2-x\log(\log(x)) &= \lim\limits_{x\rightarrow +\infty}x\log(\log(x))\left(\frac{(\log(x))^2}{x\log(\log(x))}-1\right) \\
&= +\infty(-1) \\
&= -\infty
\end{aligned}$$
Now, this was just the exponent of $e$ in the definition of our original limit. Thus:
$$\begin{aligned}
\lim\limits_{x\rightarrow +\infty}\frac{x^{\log(x)}}{\log(x)^x} &= \exp\left(\lim\limits_{x\rightarrow +\infty}(\log(x))^2-x\log(\log(x))\right)\\ 
&= \exp(-\infty)\\
 &= 0
\end{aligned}$$
So we reach the conclusion that, as $x \rightarrow +\infty$, $\log(x)^x$ grows faster than $x^{\log(x)}$.
A: For $x^{\log(x)}$ OR $(\log(x))^x$
Take t such that $x=e^t$, then
$(\log(x))^x-x^{\log(x)}=t^{e^t}-(e^t)^t=t^{e^t}-e^{t^2}=t^{e^t}-e^{e^{2\ln(t)}}$
For t>e, ($x>e^e$),
$t^{e^t}-e^{e^{2\ln(t)}}>t^{e^t}-t^{e^{2\ln(t)}}>t^{e^t}-t^{e^{t}}=0$
