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Simple math problem, exponents

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    I am working on a physics problem, and am unsure of my math at one point. The physics problem is irrelevant, as I just want to verify my math. I am in calc3 and diff eq., I am just a little fuzzy/unsure of my algebra sometimes. Could someone please tell me if the work I have below follows the proper math/algebra rules? Thanks!


    2. Relevant equations

    n/a

    3. The attempt at a solution

    Let's say I have v^-1/3 = 2P/(AN) and I want to find out what v^2/3 is equal to. In order to do this, can I just raise both sides of the equation to the -2 power so that I get this: v^-2/3 = (2P/(AN))^-2 which simplifies to this: v^-2/3 = (AN/(2P))^2 right?
     
  2. jcsd
  3. Sep 10, 2012 #2

    Simon Bridge

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    Let me rewrite all that more clearly and you can tell me if I have understood you:

    You have [tex]v^{^-\frac{1}{3}}=\frac{2P}{AN}[/tex] and you want to find [tex]v^{^+\frac{2}{3}}[/tex] ...?

    note: the rule you want is - [itex](v^a)^b = v^{ab}[/itex] so [tex](v^{^-\frac{1}{3}})^2 = v^{(^-\frac{1}{3}\times \frac{2}{1})} = v^{^-\frac{2}{3}}[/tex]... or:[tex](v^{^-\frac{1}{3}})^{^-2} = v^{(^-\frac{1}{3}\times ^- \frac{2}{1})} = v^{^+\frac{2}{3}}[/tex]
     
    Last edited: Sep 10, 2012
  4. Sep 10, 2012 #3
    Sorry for the format, I'm still trying to figure out how to use latex. Yes, that is right. I am starting out with [tex]v^{^-\frac{1}{3}}=\frac{2P}{AN}[/tex] and I wanted to use [itex](v^a)^b = v^{ab}[/itex] to find out what [tex]v^{^\frac{2}{3}}[/tex] is equal to in terms of P,A and N.
     
  5. Sep 10, 2012 #4
    Therefore, [tex]v^{^\frac{2}{3}}=(\frac{AN}{2P})^{^2}[/tex] correct?
     
  6. Sep 10, 2012 #5

    Mark44

    Staff: Mentor

    Yes.
    v-1/3 = (2P)/(AN)
    => v1/3 = (AN)/(2P)

    Now you can square both sides, getting the equation you show, above.
     
  7. Sep 10, 2012 #6
    Great! Thank you!
     
  8. Sep 10, 2012 #7

    Simon Bridge

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    OR you can raise both sides to the power of (-2) and then invert the RHS - however, it is a useful discipline to invert first: less chance of confusion.

    No worries eh?
     
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