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Logarithm question

  1. Jan 13, 2012 #1
    (-1/2)log2 + (1/2)log1 + (3/2)log4 - (3/2)log4 - (3/2)log3

    = (1/2)log[(4^3)/(2x3^3)]

    The above is from a text book. Could anyone please show me the intermediate steps between the two lines? Thank you.
     
  2. jcsd
  3. Jan 13, 2012 #2

    jedishrfu

    Staff: Mentor

    log A + log B = log AB

    N log A = log (A^N)

    similarly 2 log A + 0.5 log B = log (A^2) + log (B^0.5) = log (a^2 B^0.5)
     
  4. Jan 13, 2012 #3
    Thanks. I've been using the log laws, though I can't seem to get the same answer which is given in the book.
     
  5. Jan 13, 2012 #4

    I like Serena

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    Hi ZedCar! :smile:

    Perhaps you can show us how you applied the log laws?
     
  6. Jan 13, 2012 #5
    Question is: (-1/2)log2 + (1/2)log1 + (3/2)log4 - (3/2)log4 - (3/2)log3

    First I factored out 1/2

    1/2 [-log2 + log1 + 3log4 - 3log4 - 3log3]

    1/2 [-log(2x1) + 3log4 - 3log(4/3)]

    1/2 [-log2 + 3log4 - 3log(4/3)]

    1/2 [-log2 + 3log((4x3)/4))]

    1/2 [3log((4x3)/4) - log2]

    1/2 [3log((12x2)/4)]

    1/2 [3log(24/4)]

    1/2 [3log6]
     
  7. Jan 13, 2012 #6

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    Good!


    I'm afraid that this is not quite right.


    You have not applied the priority rules for addition and subtraction properly.
    They should be evaluated left-to-right, like this:

    1/2 [((((-log2) + log1) + 3log4) - 3log4) - 3log3]

    I've added parentheses to specify the order of evaluation.

    In particular -log2 should be evaluated as -1 x log2 = log 2-1 = log(1/2).


    Perhaps you can redo this step?
     
  8. Jan 13, 2012 #7
    Thanks very much I like Serena.

    I've been able to figure it out now! Thank you.
     
  9. Jan 13, 2012 #8
    Is this the standard way of doing these? Start from the left and work right?
     
  10. Jan 13, 2012 #9

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  11. Jan 13, 2012 #10

    I like Serena

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    Yes.
    If you have mixed additions and subtractions, you have to do them from left to right.


    However, there is an alternative, which is like this:

    1/2 [-log2 + log1 + 3log4 - 3log4 - 3log3]

    1/2 [(-1 x log2) + log1 + (3 x log4) + (-3 x log4) + (-3 x log3)]

    If you do it like this, you can add in an arbitrary order, but multiplication comes first.
     
  12. Jan 13, 2012 #11
    That's great. Thanks very much again for that!
     
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