"Find x > 3 such that
ln(x) < x^(0.1)"
How do you solve this type of problem? I plugged it into my TI-89 solver and didn't get anything. As a hint, the number is unbelievably huge.
2^60 = 1,152,921,504,606,846,976
That's one such number, but there are infinitely many more, some smaller and most larger. The proof for x=2^60 is pretty straightforward, though.
First notice that the condition
ln(x) < x^(0.1)
is equivalent to
x < e^( x^(0.1)),
so it is sufficient to show this latter property holds.
Let x = 2^60.
Then x^(0.1) = 2^6 = 64.
Since 2 < e,
2^60 < e^60
< e^64
= e^(2^6)
Quote from: RedEyes on September 01, 2010, 10:24:48 PM
2^60 = 1,152,921,504,606,846,976
That's one such number, but there are infinitely many more, some smaller and most larger. The proof for x=2^60 is pretty straightforward, though.
First notice that the condition
ln(x) < x^(0.1)
is equivalent to
x < e^( x^(0.1)),
so it is sufficient to show this latter property holds.
Let x = 2^60.
Then x^(0.1) = 2^6 = 64.
Since 2 < e,
2^60 < e^60
< e^64
= e^(2^6)
My head hurts :'(
(http://i56.tinypic.com/rhsnrm.png)
Quote from: GavinGill on September 01, 2010, 11:18:06 PM
You have a hard one for Allie? What?
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA.. :3.
number is unbelievably huge
Quote from: S on September 02, 2010, 01:31:46 AM
number is unbelievably huge
Lol, did you marquee your UB's? o_0
gavin your my hero. i lol'd :D
the heh ruined whole pic for me, sorry.