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How do u solve these kinds of problems?1/3=x^(2/3)

  1. May 29, 2005 #1
    how do u solve these kinds of problems?

    1/3=x^(2/3)
     
  2. jcsd
  3. May 29, 2005 #2

    its called algebra. you isolate the variable. in this case you would raise both sides to the reciprocal exponent.

    [tex] (\frac {1}{3})^{\frac{3}{2}} = x = \frac{1}{\sqrt{27}}[/tex]
     
  4. May 29, 2005 #3
    okay..i see...thanks....
     
  5. May 29, 2005 #4

    dextercioby

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    Gale,i think you missed a valid solution...:wink:

    Daniel.
     
  6. May 30, 2005 #5

    Ouabache

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    Perhaps Gale, you will see what Daniel means, if you break it
    down into steps

    [tex] (x^{\frac{2}{3}})^3 = ({\frac{1}{3}})^3[/tex]

    [tex] x^2 = \frac{1}{27} [/tex]

    which of course, has two solutions :wink:
     
  7. May 30, 2005 #6
    yes, i was aware thanks... but hey, if its not enough that daniel made me feel stupid for forgetting the dumb negative answer, then i feel stupider now that someone actually thought they had to explain it to me... THANKS
     
  8. May 30, 2005 #7
    Oh just in case you didnt quite catch it Gale the answers were

    [tex]\frac{\sqrt{27}}{27}, \frac{-\sqrt{27}}{27}[/tex]
     
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