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Solving Natural Logarithms

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve for x

    xln(x)-24+6ln(x)-4x=0


    My attempt at the solution

    I first moved "-24" and "-4x" to the right side of the equation yielding

    xln(x)+6ln(x)=24+4x

    I then converted the natural logarithms to exponent form and product form yielding

    ln(x)^x+ln(x)^6 = 24+4x ====> ln(x^x*x^6)=24+4x =====> ln(x^(x+6))=24+4x

    I then converted the equation in notation " e " form yielding

    e^(24+4x)= x^(x+6)

    Now I am stuck on where to proceed. I am aware that the answer has to be a positive number because the domain of a logarithm is always greater then 0.

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 1, 2011 #2

    HallsofIvy

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    That is not going to have a solution in terms of any simple algebraic function. It might have a solution in terms of the "W" function which is defined as the inverse function to [itex]f(x)= xe^x[/itex].
     
  4. Nov 1, 2011 #3

    SteamKing

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    You were on the right track when you collected like terms:
    xln(x)+6ln(x)=24+4x

    If you factor the LHS further, notice:
    ln(x) (x+6)=24+4x

    Can you figure out what x must be now? (Sorry, Mr. Lambert. No soup for you today.)
     
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