[Racenata]

"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.

[Santa]

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)

[zomniethe4]

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 

[GavinGill]

[Xrow]

You have a hard one for Allie? What?

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA.. :3.

[S]

number is unbelievably huge

[Xrow]

number is unbelievably huge

Lol, did you marquee your UB's? o_0

[Corr]

This man's a winner.

[7ehNoob]

gavin your my hero. i lol'd

[Yz]

the heh ruined whole pic for me, sorry.