Trinity Cambridge Admissions Test

I feel like I'm missing something quite important in this question.

${{2^{2}}^{2}}^{n}$ is equal to $16^{n}$, and we can clearly see that if $n=1, 16<100^{100}$, so then as $n$ gets large $g(n)={100^{100}}^{n}$ is greater than $f(n)$ ?

compare the log of the log of each expression

I don't really understand how the log of a log works.

I feel like I'm missing something quite important in this question.

${{2^{2}}^{2}}^{n}$ is equal to $16^{n}$, and we can clearly see that if $n=1, 16<100^{100}$, so then as $n$ gets large $g(n)={100^{100}}^{n}$ is greater than $f(n)$ ?

Here's the clever way I did it, consider $h(n)=256^{256^n}$, which is basically just logs I guess.

I don't really see how i can relate the result from this to the functions given in the question, though?

So clearly, $h(n)>g(n) \forall n$. We write
$[br]h(n)=256^{256^n}[br]=2^{8*256^n}[br]=2^{2^3*2^{8n}}[br]=2^{2^{8n+3}}[br]$
and the result should be clear from this.

So clearly, $h(n)>g(n) \forall n$. We write
$[br]h(n)=256^{256^n}[br]=2^{8*256^n}[br]=2^{2^3*2^{8n}}[br]=2^{2^{8n+3}}[br]$
and the result should be clear from this.

me?

Another user to add to my "refuse to help" list ...

No it isn't.

With a "power tower" you calculate from "top to bottom". E.g.

$2^{2^{2^4}} = 2^{2^{16}} = 2^{65536}$, which is much larger than 16^4.

That's a little rude.

I'm still working on your way to be honest, i'm just finding it hard to picture how to get my head around it without using a calculator. Just because I haven't replied yet, it's not very fair to say that!

Ah I only considered when $n=1$ which was silly as the question says as it gets larger

Do these type of numbers come up much in mathematics in general? (the "power towers"

I stopped my own work, wrote an almost complete solution, scanned it, posted it, and you failed even to acknowledge my post, let alone thank me...
Sorry that I am rude, how bad of me ...
I guess I am invisible.

Well I apologise for not immediately responding, but usually if it's maths I try to work through the advice first for a while until I either get the answer or at least get to the point where I can ask a question that's not stupid~ It's bad enough that I have to ask for help on questions that some people seem to find so easy. I'd rather have replied once I got it from your help, than to have said "Thankyou, but I still can't see it.."

BTW I did try to rep you when you posted it but it said I've rep'd you too much lately

q1
http://www.bmoc.maths.org/home/bmo1-2015.pdf