Jump to content

  • Log In with Google      Sign In   
  • Create Account

equivalent values


Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

  • You cannot reply to this topic
3 replies to this topic

#1 outRider   Members   -  Reputation: 852

Like
Likes
Like

Posted 05 October 2001 - 03:50 PM

I find myself having to find alternate representations of different types of terms, basically fractional and exponential representations, to solve certain problems. I can''t for the life of me remember what equals what. For example I know that 1 / x^4 = x^-4 and 3x^-4 = 3 / x^4 Here''s the ones I''m havin trouble with, I need the fractional/exponential representation for each. I know that I might actually list both representations. aX^n = aX^-n = a / bX^n = a / bX^-n = -a / bX^n = -a / bX^-n = X^1/2 = sqrt(X) ? X^-a/b = a / X^b/c = I know thats a lot, so if anyone has even just a few examples that would work in any case then that might be better. Thanks in advance to anyone who can help. ------------ - outRider -

Sponsor:

#2 Dean Harding   Members   -  Reputation: 546

Like
Likes
Like

Posted 05 October 2001 - 05:03 PM

All you need to remember is that a negative exponent means you invert the fraction. ie:

x-n = 1/xn

or

1/x-n = xn

Also, the denominator of an exponent is the same as the order of the root of the number, for example:

x1/2 = sqrt(x)
x1/3 = cuberoot(x)
x3/2 = sqrt(x3)
x-3/2 = sqrt(x-3) = sqrt(1/x3)

Hope this helps!


codeka.com - Just click it.

#3 Scarab0   Members   -  Reputation: 122

Like
Likes
Like

Posted 05 October 2001 - 08:57 PM

In general:
xa/b = b-root(xa)

#4 outRider   Members   -  Reputation: 852

Like
Likes
Like

Posted 07 October 2001 - 10:36 AM

Thank you so kindly.

------------
- outRider -




Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.



PARTNERS