x^(2/3) (1 Viewer)

CrashOveride

Active Member
Joined
Feb 18, 2004
Messages
1,488
Location
Havana
Gender
Undisclosed
HSC
2006
Originally posted by Estel
hmm

suppose x = a/b where a/b is negative
y = x^x
= (a/b)^(a/b)
= 1/[(a/b)^(|a/b|)]
= 1/[(|b|rt)(a/b)^|a|]
hence if a is odd b must be odd
and if a is even b can be either.

so x = -0.4 = -2/5 should work then.
Where's the fault in my logic?

Edit: I see. Cursed calculator.
Your last line should read:
y= 1/[(|b|rt(a/b))^|a|]

Maybe clearer with lil example:

x<sup>2/3</sup> = x<sup>1/3<sup>2</sup></sup>

So its the cube root of x, squared
 
Last edited:

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
So you take the root first.

That elucidates everything. :)
Thanks.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top