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Solving equation with negative exponents

  1. Jun 17, 2009 #1
    1. The problem statement, all variables and given/known data

    2x^(-1/3)-9x^(-1/6)= -10

    2. Relevant equations

    ?

    3. The attempt at a solution
    I have tried to factor out x^(-1/6)
    x^(-1/6) (2x-9)= -10
    I'm not sure thats even right
    I have also converted to fractions
    1/2x^(1/3)-1/9x^(1/6)= -10
    I'm not sure which route to go or if either is right???
     
  2. jcsd
  3. Jun 17, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    If you write it this way, can you see what you can do to put the left hand side (LHS) over a common denominator in order to proceed?

    [tex]\frac{2}{x^{1/3}} - \frac{9}{x^{1/6}} = -10[/tex]
     
  4. Jun 17, 2009 #3

    Mark44

    Staff: Mentor

    And as a further hint, x1/3 = (x1/6)2, so with the right substitution, your equation is quadratic in form.
     
  5. Jun 18, 2009 #4
    Yeah always try to notice when you can turn an equation into a quadratic eg.

    [itex] x + x^{-1} + A = 0 \Rightarrow x^2+1+Ax= 0[/itex]


    [itex] e^x + e^{-x} + A = 0 \Rightarrow (e^{x})^2 +1+ Ae^x = 0[/itex]


    [itex] \cot(x) + \tan(x) + A = 0 \Rightarrow 1 + \tan^2 x +A\tan x= 0[/itex] etc.

    substitutions can be helpful aswell, like substitute e^x for y or something.
     
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