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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.

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  • $\begingroup$ Is your second expression $\log (x^x)$ or $(\log x)^x$? $\endgroup$
    – njguliyev
    Oct 19, 2013 at 11:12
  • $\begingroup$ it (logx)^X , but I thought there are the same, please correct me if I'm wrong. $\endgroup$
    – user836026
    Oct 19, 2013 at 11:13
  • $\begingroup$ No, $\log (x^x) = x\cdot\log x$. $\endgroup$
    – njguliyev
    Oct 19, 2013 at 11:14
  • $\begingroup$ sorry I will edit question $\endgroup$
    – user836026
    Oct 19, 2013 at 11:15
  • $\begingroup$ Alternatively, you could take the $\log$ of both functions and compare $(\log x)^2$ with $x\log\log x$. $\endgroup$ Jun 12, 2020 at 9:17

3 Answers 3

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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 $x^{\log(x)}$.

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Hint: $\dfrac{\ln y}{y}$ is decreasing for $y > e$.

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

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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, ($x>e^e$), $t^{e^t}-e^{e^{2\ln(t)}}>t^{e^t}-t^{e^{2\ln(t)}}>t^{e^t}-t^{e^{t}}=0$

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