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Tricky equation

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

    [tex]D(e^{-2ax}-2e^{-ax})-E=0[/tex]

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

    HallsofIvy

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    Staff Emeritus
    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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