# Showing that $2^{2^{2^x}}<100^{100^x}$ for large $x$

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.

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

## 2 Answers

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

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

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

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

$$100^{100^x}= 2^{[\log_2{100}]*100^x}$$. 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.