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Natural Logarithm Manupulations

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]xln(2x+1)-x+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)-x[/tex]

    2. Relevant equations
    [tex]ln(x^a) = aln(x), ln(xy) = ln(x) + ln(y), ln(\frac{x}{y}) = ln(x) - ln(y)[/tex]

    3. The attempt at a solution

    I have no idea how you can go from [itex] xln(2x+1)-x+\frac{1}{2}ln(2x+1)[/itex] to [itex]\frac{1}{2}(2x+1)ln(2x+1)-x[/itex] could someone point me in the right direction?

    I know both sides have the -x term, so the only change takes place in [itex]xln(2x+1)+\frac{1}{2}ln(2x+1) = \frac{1}{2}(2x+1)ln(2x+1)[/itex]
    Last edited: Mar 19, 2007
  2. jcsd
  3. Mar 19, 2007 #2
    factor out the ln(2x+1)
  4. Mar 19, 2007 #3
    :surprised wow I can't believe I didn't see that, thanks so much I get it now
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