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Ln(9/4) + ln (16/9) - ln (3/1)

  1. Dec 31, 2013 #1

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    1. The problem statement, all variables and given/known data

    If you have, for example, 2 + 4 - 1, you can get the answer (5), by doing both:

    = 2 + (4 - 1)

    and,

    = (2 + 4) - 1


    But the same logic does not work with logs: to get the right answer (4/3) here you must do:

    =(ln(9/4) + ln (16/9)) - ln (3/1)

    and NOT:

    ln(9/4) + (ln (16/9) - ln (3/1))


    Why exactly is this so?



    2. Relevant equations

    ln (9/4) + ln (16/9) - ln (3/1)

    2. Attempt at a solution

    (ln (9/4) + ln (16/9)) - ln (3/1)

    = (ln (4) - ln (3))
    = ln (4/3) = correct

    (ln (9/4) + (ln (16/9)) - ln (3/1))

    = ln (9/4) + ln (16/27)
    = ln (144/108) = wrong
     
    Last edited: Dec 31, 2013
  2. jcsd
  3. Dec 31, 2013 #2
    What does that reduce to in lowest terms?
     
  4. Dec 31, 2013 #3

    939

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    Ahhhh, such a stupid mistake. Thanks.
     
  5. Dec 31, 2013 #4
    It happens to everybody, you're welcome.

    Rather than simplify at the end you can do it earlier by recognizing that 9 divides into 27 evenly and 4 divides into 16 evenly: $$\frac{9}{4} \cdot \frac{16}{27} = \frac{16}{4} \cdot \frac{9}{27}$$.
     
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