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From: Gc on 10 Aug 2010 06:29
On 10 elo, 12:28, Rob Johnson <r...(a)trash.whim.org> wrote:
> In article <dd0464f6-0c3f-4b6c-b3cf-8895e792b...(a)f42g2000yqn.googlegroups..com>,
>
> Gc <gcut...(a)hotmail.com> wrote:
> >On 8 elo, 20:39, Shin Dongjun <acam...(a)gmail.com> wrote:
> >> Prove (log3)^2  > (log4)^3.
>
> >> (Don't use an approximation like log3 = 0.4771)
>
> >> log is common logarithms that is, base of log is 10.
>
> >log(3)^2 = log(12/4)^2 = (log(12) - log(4))^2 = log(12)^2 -
> >2log(12)log(4) + log(4)^2 > log(4)^2, because log(12)log(12) >
> >2log(12)log(4) = log(12)log(8).
>
> 2 log(4) = log(16) not log(8).  A big tip-off that something is wrong
> is that you've just proven that log(3)^2 > log(4)^2, which is false.

Yeah, sorry, I noticed that soon. I hope that in the usenet you could
edit your posts :( I try to think more the next time when I post. Now
I have to stop posting here for a while. I got other stuff to do.
From: Shin Dongjun on 13 Aug 2010 10:57
On 8¿ù9ÀÏ, ¿ÀÀü2½Ã39ºÐ, Shin Dongjun <acam...(a)gmail..com> wrote:

Thanks for replies, guys...

I read all. I really appreciate you for trying to solve it.

But I think I still don't have a good solution.

Of course I also, like you all, solved it by brute force calculation.

But I believe someone has a beautiful solution...

If you solve it nicely, please post it anytime.^^

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