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Homework Help: Math help

  1. Feb 12, 2005 #1
    How do I make y the subject?
    Then what?
  2. jcsd
  3. Feb 12, 2005 #2


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    y >= 0
    [tex]y = \sqrt[\frac{2}{3}]{\frac{x}{3}} = \sqrt{(\frac{x}{3})^{3}} [/tex]
    Hope it help,
    Viet Dao,
  4. Feb 12, 2005 #3
    How did you go from [tex]\sqrt[\frac{2}{3}]{\frac{x}{3}}[/tex] to [tex]\sqrt{(\frac{x}{3})^{3}}[/tex]
  5. Feb 12, 2005 #4
    Using the law of indicies that says a^mn = (a^m)^n.

    y^(2/3) = x/3

    Cube both sides: y^2 = (x/3)^3

    Now square root both sides: y = (x/3)^3/2 = [(x/3)^3]^1/2, which is what you have (I can't use LaTeX properly, oops).
  6. Feb 12, 2005 #5


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    That root in the LHS is another way of writing
    [tex] (\frac{x}{3})^{\frac{3}{2}}=[(\frac{x}{3})^{3}]^{\frac{1}{2}}=\sqrt{(\frac{x}{3})^{3}} [/tex]

  7. Feb 12, 2005 #6
    Oh ya! Forgot about that. :rolleyes:
    Thanks for the help guys!
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