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Expanding ln isn't working for me

  1. Sep 14, 2013 #1
    original equation: ln(0.2048x[itex]\frac{5}{2}[/itex])= -[itex]\frac{5}{2}[/itex]

    when i just raise this to e i get..

    0.2048T[itex]\frac{5}{2}[/itex]= -[itex]\frac{5}{2}[/itex]

    and then i can solve for x and get the right answer. but when i expand the logs..

    ln(0.2048) + [itex]\frac{5}{2}[/itex]ln(x)= -[itex]\frac{5}{2}[/itex]

    0.2048+[itex]\frac{5}{2}[/itex]x=e-[itex]\frac{5}{2}[/itex]

    then i try to solve for x i don't get the right answer. what am i doing wrong?
     
  2. jcsd
  3. Sep 14, 2013 #2

    Pythagorean

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    Gold Member

    you can't do that with a sum of ln's. Taking the exponent of both sides after expansion does not cancel out the ln in each term. You would have e^(lnA + lnB) and that does not equal A+B.
     
  4. Sep 14, 2013 #3
    oops, lol thanks
     
  5. Sep 14, 2013 #4

    Pythagorean

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    no problem
     
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