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I wish to show that $2^{2^{2^x}}<100^{100^x}$ for $x$ sufficiently large. I have taken logs (base 10) of both sides to get $2^{(2^x-1)}\log_{10} 2$ and $100^x$. It is not immediately clear how I can prove that the second term here is larger than the first.

Any help appreciated.

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    $\begingroup$ what if you take the log again? on the left side you get up to constant $2^x$ and on the right side $x$... $\endgroup$ – ALG Oct 31 '18 at 23:10
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    $\begingroup$ Are you sure you have the inequality pointing in the correct direction? $\endgroup$ – Barry Cipra Oct 31 '18 at 23:13
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    $\begingroup$ Actually for sufficiently large $x$ it does not hold. $\endgroup$ – André Porto Oct 31 '18 at 23:19
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HINT

Since $\log$ function is strictly increasing we have

$$2^{2^{2^x}}<100^{100^x}\iff \log_{10}\left(2^{2^{2^x}}\right)<\log_{10}\left(100^{100^x}\right) \iff 2^{2^x}\log_{10}2<2\cdot 100^x$$

then again

$$2^{2^x}\log_{10}2<2\cdot 100^x \iff \log_{10}\left(2^{2^x}\log_{10}2\right)<\log_{10}\left(2\cdot 100^x\right)$$

$$2^x\log_{10}2 + \log_{10}(\log_{10}2)<\log_{10}2+2x$$

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    $\begingroup$ Which prove that $LHS \gt RHS$ for $x\gt 6$ (and also for some number in $[5,6])$. The statement if not true. $\endgroup$ – Piquito Oct 31 '18 at 23:52
  • $\begingroup$ @Piquito Nice additional hint! Thanks a lot my dear friend! Cheers $\endgroup$ – user Oct 31 '18 at 23:54
  • $\begingroup$ You are welcome. Regards. $\endgroup$ – Piquito Nov 1 '18 at 0:01
  • $\begingroup$ How would you prove that LHS > RHS for $x > 6$? Would induction be a wise approach? $\endgroup$ – wrb98 Nov 1 '18 at 1:34
  • $\begingroup$ @wrb98 We can use induction. $\endgroup$ – user Nov 1 '18 at 8:00
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$100^{100^x}= 2^{[\log_2{100}]*100^x} $. Let $k = \log_2{100}$. ($6< k < 7$)

$100^{100^x} = 2^{k*2^{kx}}= 2^{2^{\log_2k}*2^{kx}}$. Let $m = \log_2 k$. ($2< m < 3$)

$100^{100^x} = 2^{2^m*2^{kx}}=2^{2^{m + kx}}$

Now $2^{2^{m+kx}}< 2^{2^{2^x}} \iff m + kx < 2^x$ for sufficiently large $x$.

Which should be enough to be convincing.

But if not:

If $x > m$ then $m + kx < x +kx = (k+1)x< 8x=2^3*x$.

And if we can show $x < 2^{x-3}$ for sufficiently large $x$ we are done.

which... of course it is.

$(x)' = 1$ and $(2^{x-3})'= \ln 2* 2^{x-3}> 1$ for all $x > ...$ well $x > 3 + \log_2 (\frac 1{\ln 2})= 3 + \frac {\ln \frac 1{\ln 2}}{\ln 2}\approx 3.52$. then $2^{x-3}$ is increasing faster than $x$. So at $x = 6>3 + \frac {\ln \frac 1{\ln 2}}{\ln 2}$ we have $x = 6 < 8 = 2^{x-3}$ and $2^{x-3}$ is increasing faster than $x$ so for $x \ge 6$ we will have our result.

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