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Natural Logarithms

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data

    lnx+ ln(x-1) = 1

    solve each equation for x

    2. Relevant equations

    ln(e^x) = x
    e^lnx = x

    3. The attempt at a solution

    x + (x-1) = e^1 [==> using ln(e^x) = x]
    from this point on, I am stuck because I am having trouble isolating x because of the x that is in the brackets.

    -Thanks in advance
     
  2. jcsd
  3. Sep 20, 2009 #2
    Remember the logarithm property
    ln(a) + ln(b) = ln(ab)
     
  4. Sep 20, 2009 #3
    Bohrok,

    Thank you. For some reason I always assumed that the logarithm laws such as log(x/y)= log x-logy etc, could not be applied to natural logarithms. I guess questions like that never really came up.

    So using ln(ab) = lna + lnb

    I get,

    lnx + ln(x-1) = 1
    ln(x2-x) = 1

    so e1 = x2-x

    Am I missing a step? I still cannot isolate x to solve for it. I think I may be jumping in a little too early for the cancellation rule.

    -Thanks in advance
     
  5. Sep 20, 2009 #4
    The properties of logarithms work with any positive number base (I believe), at least with most numbers you come across, like 10 and e.

    What you have now is a quadratic equation with an x2 term, so use the quadratic formula after you set the equation equal to 0.
     
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