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Natural Logarithmic Function

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve ln(2x+1)=2-ln(x) for x.

    2. Relevant equations

    3. The attempt at a solution

    e^(ln(2x+1)) = e^((2-ln(x))

    2x + 1 = e^(2/x)

    2x^2 + x = e^2

    2x^2+x-e^2 = 0

    At this point, I know you're supposed to use the quadratic equation. But, my problem is with how to treat e^2. Should I multiply it out in the discriminant where c = e^2? or just e?

    Thanks for any help.
  2. jcsd
  3. Jun 9, 2008 #2


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    e^2 is just a number... you can call it c if you like.
  4. Jun 9, 2008 #3


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    You can either just calculate it out (it's only a number!) or include it in the quadratic formula and quote your answers in terms of e.
    It depends wether this is for home work or real work!
  5. Jun 9, 2008 #4
    There is an error in the second part

    e^(ln(2x+1)) = e^((2-ln(x))

    2x + 1 = e^(2/x)

    These are NOT equivalent. Remember that you can only turn logarithmic expressions
    into quotients of logarithms in a case like this: ln z - ln x = ln (z/x)

    Try again using that info.
  6. Jun 9, 2008 #5


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    I suspect that was a typo because in the next line he has
    2x^2+ x= e^2 which is correct.

    Added Later: In fact, I would have been inclined to solve the problem as follows:
    ln(2x+1)=2-ln(x) so, adding ln(x) to both sides, ln(2x+1)+ ln(x)= ln(x(2x+1))= 2.
    Now take the exponetial of both sides: x(2x+1)= 2x2+ x= e2 and you can solve that quadratic equation for x.
    Last edited by a moderator: Jun 10, 2008
  7. Jun 9, 2008 #6


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    BrendanH, Rossinole meant (e^2)/x, not e^(2/x). The next line in the derivation,
    2x^2 + x = e^2
    is correct.

    edit added:
    Halls beat me to it.
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