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Tricky equation

  1. Nov 27, 2006 #1
    How do I solve


    for x?
  2. jcsd
  3. Nov 27, 2006 #2


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    Science Advisor

    D and E are constants?

    [itex]e^{-2ax}= (e^{ax})^2[/itex]. Let [itex]y= e^{-ax}[/itex] and your equation becomes [itex]D(y^2- 2y)- E= 0[/itex]. Solve that equation, using the quadratic formula perhaps, and then [itex]x= -ln(y)/a[/itex].
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