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Homework Help: Solve Logx^2

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Q: Logx = 1 + 6/logx

    my attempt so far: Log x^2 = 7

    where do i go from here?
  2. jcsd
  3. Feb 5, 2010 #2


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    Homework Helper

    Re: Logx^2

    How did you obtain this result? Try to multiply the whole equation with logx. You'll arrive at a quadratic equation. Try to use a substitution then.
  4. Feb 5, 2010 #3


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    Science Advisor

    Re: Logx^2

    Is this [itex]log x= 1+ \frac{6}{log x}[/itex] or [itex]log x= \frac{1+ 6}{log x}[/itex]?

    I suspect it is the former because otherwise it would just be written as 7/ log x. But then multiplying log x= 1+ (6/log x) by log x you get (log x)^2= log x+ 6. Let y= log x and that becomes y^2= y+ 6. Solve that quadratic equation for y, then solve log x= y for x.
  5. Feb 5, 2010 #4
    Re: Logx^2

    yeah it's the last one.

    so then i end up with:

    y^2-y-6 = 0

    x1 = 3 , and x2 = -2

    is that the answer?
  6. Feb 5, 2010 #5


    Staff: Mentor

    Re: Logx^2

    You mean the first one, which is logx = 1 + (6/logx).
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