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Simple Indicies Question

  1. Dec 3, 2011 #1
    Simple Indicies Question [SOLVED]

    1. The problem statement, all variables and given/known data
    I'm having a little problem with indicies, I know it's simple for someone with a lot of question.

    So I'm wondering what to do first in this question, brackets or should I multiply out the fractions inside the brackets?

    I have to simplify it that's all :)

    2. Relevant equations


    3. The attempt at a solution

    I don't have an attempt yet sorry :(
    Last edited: Dec 3, 2011
  2. jcsd
  3. Dec 3, 2011 #2
    I'd start by working out the inside of the brackets first. I think that'll be the easiest.
  4. Dec 3, 2011 #3
    Thanks for the reply, so when I worked it out and simplified it first I got (x^2.5 x^2 x^-3)^2 which would equal to something like x^3, but in derive the answer is x^17..

    What am I doing wrong :(?
  5. Dec 3, 2011 #4
    That X-3 isn't correct, is it??
  6. Dec 3, 2011 #5
    Well I thought it was... What else could it be? I wish I had a good course tutor in College, I literally teach myself almost everything! *try*
  7. Dec 3, 2011 #6

    Ray Vickson

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    You have [tex]F = \left( \frac{X^4 X^5 X}{X^{1.5} X^3 X^{-3}}\right)^2 . [/tex] The first step is to simplify the quantity inside the bracket, to obtain [itex] F = (X^a)^2. [/itex] So, the first order of business is to figure out what is 'a' in the following:
    [tex] \frac{X^4 X^5 X}{X^{1.5} X^3 X^{-3}} = X^a. [/tex] After that, the rest is easy: [itex] (X^a)^2 = X^{2a} . [/itex]

  8. Dec 3, 2011 #7
    Sorry, common mistake, so it would be X^3?
  9. Dec 3, 2011 #8
    Here's what I got... That previous post is very complicated :eek:

  10. Dec 3, 2011 #9


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    You need to slowly apply these rules, you keep making mistakes.

    [tex]a^b\cdot a^c=a^{b+c}[/tex]


  11. Dec 3, 2011 #10
    How is [itex]X^{-3}[/itex] defined?? What is [itex]\frac{X}{X^{-3}}[/itex]??

    Are you aware of the identity [itex]\frac{a^n}{a^m}=a^{n-m}[/itex]??
  12. Dec 3, 2011 #11
    Thanks guys I have solved this problem now :) I will keep going over and over until I get it perfect.
  13. Dec 3, 2011 #12


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    Could you show us just to be sure? Because two wrongs can sometimes accidentally make a right :tongue:

    And assuming you used the formulae correctly, just a tip, it'll probably be easier if you simplify the numerator first, then the denominator, then apply the quotient rule.
  14. Dec 3, 2011 #13
    What I did first was simplify the insides of the brackets. To do this I applied the 2nd law of indicies and take away the denominator from the numerator for example, first of all I got x^2.5 because 4-1.5 = Positive 2.5... Then the same for the next one in the brackets.

    Now the last fraction in the equation at first I forgot the rule of two the same signs make positive and the opposite signs make a negative. So x-(-3) would be equal to x^3.

    Then once I got all of them, I added them up to form (x^7.5)^2

  15. Dec 3, 2011 #14

    Ray Vickson

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    OK, that works, but you still have made some errors. However, what people are suggesting is that you do it more systematically, by simplifying the numerator and denominator separately:
    [tex] \mbox{numerator} = X^4 X^5 X = X^{4+5+1} = X^{10}[/tex] and
    [tex] \mbox{denominator} = X^{1.5} X^3 X^{-3} = X^{1.5 + 3 - 3} = X^{1.5}, [/tex] to get [tex] \mbox{ratio} = \frac{\mbox{numerator}}{\mbox{denominator}} = \frac{X^{10}}{X^{1.5}} = X^{10 - 1.5} = X^{8.5}. [/tex] There is less chance of making an error when you do it this way.

  16. Dec 3, 2011 #15
    Oh right I see what you mean! I told you my Tutor was rubish :P I'll get into that routine then, Thank you! :)
  17. Dec 4, 2011 #16


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    How did you get from (x7.5)2=x17?
  18. Dec 4, 2011 #17
    Ah Sorry I must of confused my self somewhere... I meant x^8.5 at least thats what I have on paper..
  19. Dec 4, 2011 #18


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    Ahh ok just a typo then, because you did it twice :wink:
  20. Dec 4, 2011 #19
    Lmao not sure why I did it twice, I was very tired that night :P
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