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So, I have got this problem in which one is asked to find the greater one between $10^{30!}$ or $10^{30}!$.

Taking $\log$ both sides leads me nowhere.

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  • 1
    $\begingroup$ $({}{}10^{30})!$ $\endgroup$ – Lord Shark the Unknown Jun 19 '18 at 18:14
  • $\begingroup$ You may compare the max no. of digits of first one $10^{30!}$ with min. of digits in $10^{30}!$. Second one will be bigger as it will have more digits. $\endgroup$ – Love Invariants Jun 19 '18 at 18:16
  • $\begingroup$ Put in Stirling's approximation and then take the log. $\endgroup$ – Ross Millikan Jun 19 '18 at 18:16
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    $\begingroup$ Interestingly $f(x)=10^{x!} $ and $g(x)=10^x!$ we have for $f(27)<g(27)$ but $f(28)>g(28)$ . Nice problem! $\endgroup$ – Gottfried Helms Jun 20 '18 at 9:59
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You'll want Stirling's approximation here: $\ln n! \approx n \ln n - n$. Here $\ln$ is the natural log.

So $\ln (10^{30})! \approx 10^{30} \ln 10^{30} - 10^{30}$. Now $\ln 10^{30} = 30 \ln 10 < 30 \times 3 < 100$, and so $\ln (10^{30})! < 10^{32}$.

On the other hand, $\ln 10^{30!} = 30! \ln 10$. You can verify that $30! > 10^{32}$. and so $\ln 10^{30!} > 10^{32}$.

Thus $10^{30!}$ is the larger of the two.

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  • $\begingroup$ I was wondering why nobody was just 2x checking... $\endgroup$ – Antoni Parellada Jun 19 '18 at 18:23
  • $\begingroup$ How far away from the origin is the approximation relatable though? Or is that not of significance here? $\endgroup$ – gen-z ready to perish Jun 19 '18 at 18:58
  • $\begingroup$ @MathAsFun: Imagine you are asked for solve this problem in an exam. What do you do? Wolfram is a good tool for check the answer but it isn't the answer. Of course this question doesn't worth for me, it is just curious and wolfram is enough but... For other people this question is important. $\endgroup$ – Dog_69 Jun 19 '18 at 21:35
  • $\begingroup$ @Dog_69 You're missing the point of my comment, but it's unimportant. Best wishes! $\endgroup$ – Antoni Parellada Jun 19 '18 at 21:38
  • $\begingroup$ @MathAsFun: Maybe. In this case sorry. $\endgroup$ – Dog_69 Jun 19 '18 at 21:40
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$x! < x^x\\ 10^{30}! < (10^{30})^{10^{30}}$

While $10^{30!} = (10^{30})^{29!}$

Since $10^{30} > 1$, if $29! > 10^{30}$ (which it is) then $10^{30!} > 10^{30}!$

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  • $\begingroup$ @hardmath Really? I thought what I had was pretty easy to follow. $\endgroup$ – Doug M Jun 19 '18 at 20:07
  • $\begingroup$ In my opinion the proof is clear. However, I would like to see a proof for the inequality $29!>10^{30}$, because it's not obvious. I mean, you are proving that $10^{30!}>10^{30}!$, so to do a fantastic answer you should prove the other because isn't obvious (at least for me). $\endgroup$ – Dog_69 Jun 19 '18 at 21:31
  • $\begingroup$ @Dog_69: It's rather tight in that $10^{30} \lt 29! \lt 10^{31}$. $\endgroup$ – hardmath Jun 19 '18 at 23:12
  • $\begingroup$ Using Pari/GP and the log- and lngamma-function I get a critical point between $27$ and $28$ (in place of $30$): [log(10)*27!;lngamma(10^27)] --> [2.50725484761 E28; 6.11697975108 E28] and [log(10)*28!;lngamma(10^28)] --> [7.02031357331 E29; 6.34723826038 E29] $\endgroup$ – Gottfried Helms Jun 20 '18 at 10:06

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