Fractals: Problem with understanding log

Discussion in 'Homework Help' started by zulfi100, May 9, 2015.

  1. zulfi100

    Thread Starter Member

    Jun 7, 2012
    320
    0
    Hi,
    I am studying fractals. It uses natuaral log. I cant understand how the author developed the second equation from the first. The first equation is:
    Zufi100_fractal-eq1.png
    And the second equation is:
    Zufi100_fractal-eq2.png
    Can some body please guide me how to obtain the second equation. I tried the following:
    ln ns^D = ln (1)
    ln(n) + ln(s^D) = 1
    ln(n) + Dln(s) = 1
    Dln(s) = -ln(n)
    D = -ln(n)/ ln(s)

    Some body please guide me.

    Zulfi.

    Moderators note: removed white space from picture and changed contrast and brightness
     
    Last edited by a moderator: May 9, 2015
  2. WBahn

    Moderator

    Mar 31, 2012
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    What is the ln(1/s) in terms of the ln(s)?
     
  3. zulfi100

    Thread Starter Member

    Jun 7, 2012
    320
    0
    Hi,
    Thanks for your response. I can tell you about ln(1/s). Its :
    ln(1) - ln(s).
    Please guide me how to obtain 2nd equation from 1st equation as i posted in #1.
    Zulfi.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    4,789
    And what is ln(1)?
     
  5. zulfi100

    Thread Starter Member

    Jun 7, 2012
    320
    0
    Its 1. Also tell me if the above is correct or not? If its correct then it means that if we convert ln(s) into ln(1/s) we will get a negative sign. So now we have :

    ln ns^D = ln (1)
    ln(n) + ln(s^D) = 1
    ln(n) + Dln(s) = 1
    Dln(s) = -ln(n)
    D = -ln(n)/ ln(s)
    D= ln(n)/ln(1/s)

    Is the above correct?

    Zulfi.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,737
    4,789
    If the ln(1) = 1, then that means that e^1 = 1. Is that correct?

    What must the base, e, be raised to in order to get a result of 1?

    In your work, how do you justify the following step:

    ln(n) + Dln(s) = 1
    Dln(s) = -ln(n)

    Where did the 1 go?
     
  7. zulfi100

    Thread Starter Member

    Jun 7, 2012
    320
    0
    Hi,
    Okay. I got my mistake.
    ns^D = 1
    ln(ns^D) = ln(1)
    ln n + ln s^D = ln (1)
    ln s^D = ln (1) - ln n
    D = ( ln 1 - ln n) / ln s
    D = (0 - ln n)/ln s
    D= ln n / ln (1/ s)
    Is this correct now???
    I got the answer from google : ln (1) is actually zero.

    Please guide me.

    Zulfi.
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,737
    4,789
    That you had to Google what the ln(1) is does not bode well for you. You need to understand these fundamental concepts. Not memorize them, not be able to Google them, but to understand them.

    Your work is correct now, but rather convoluted. It can be presented much more cleanly as

    n·s^D = 1
    ln (n·s^D) = ln(1)
    ln(n) + D·ln(s) = 0
    D·ln(s) = -ln(n)
    D = - ln(n)/ln(s)
    D = ln(n)/ln(1/s) = ln(1/n)/ln(s)

    Unless (1/s) has some particularly meaning, I would probably leave it as the next to last line.
     
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