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