1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Why is this?
  1. Why not a topology? (Replies: 2)

  2. Why is it zero ? (Replies: 3)

  3. Why is this? (Replies: 3)

Loading...