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Why is this?

  1. Dec 13, 2006 #1
    Why is it that ln(x^2+1) = ln (abs: 2y-3) can be transformed into x^2+1 = 2y-3 , this according to an example in my book.

    What I don't understand is why you can eliminate the absolute sign.
     
  2. jcsd
  3. Dec 13, 2006 #2
    Huh ? The argument of the ln function always needs to be positive by definition : if ln(A) = B then exp(B) = A.

    marlon
     
  4. Dec 13, 2006 #3
    http://www.math.ntnu.no/emner/kode/SIF5003/gamle-eks/TMA4100_2006_08_18_lf.pdf [Broken]

    It's task 3 here. You won't understand the language, but hopefully understand the maths :smile:
     
    Last edited by a moderator: May 2, 2017
  5. Dec 13, 2006 #4
    That's Norwegian, right ?

    yep, they write the absolute sign because the argument of the ln function NEEDS to be positive. So in that case, you only consider the y values for which 2y-3 > 0 --> y>1.5

    marlon
     
  6. Dec 13, 2006 #5
    I don't really understand why it needs to be positive. Is it not possible to say that e^(ln(abs: 2y-3)) = (abs: 2y-3) ?

    And yes, it's Norwegian. Did you guess that without checking my profile?
     
    Last edited: Dec 13, 2006
  7. Dec 13, 2006 #6
    e^(ln(abs: 2y-3)) = (abs: 2y-3) : YES THAT IS CORRECT.

    Now, if you are using the absolute value function like |x| and you know that x > 0 (because of the argument of ln), the definition of |.| says that |x| = x.

    In short : definition for |.|

    |x| = x if x > 0
    |x| =-x if x < 0

    So |2y-3| = 2y-3 IF 2y-3 > 0

    Finally, let me repeat why the argument (2y-3) of the ln function is positive BY DEFINITION:

    ln(A) = B <--> A = exp(B)

    But exp(B) is always positive so A must always be positive.

    Keep in mind that exp(B) = [tex]e^B = (2.7)^B[/tex]

    When you exponentiate 2.7, it can never become a negative number.



    Yes, my native language is Dutch and some of the words look very much alike.

    marlon
     
    Last edited: Dec 13, 2006
  8. Dec 13, 2006 #7

    HallsofIvy

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    The citation you give doesn't say that! It says "if ln(x2+ 1)= ln(|2y- 3|)+ C1, then x2+ 1= C1(2y-3)".

    You don't need the "| |" because the C1 can be chosen positive or negative.
     
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