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Homework Help: Logarithm overkill!

  1. Jan 23, 2008 #1
    [SOLVED] Logarithm overkill!

    Hello again!
    I have been working on this log, and the longer I work on it, the more confused I get! Here's the problem:
    Find the exact value for:

    [tex]ln(ln[e^{e^{5}}[/tex]])

    ----
    Here's what I've tried so far:

    [tex]e(ln[e^{e^5}}[/tex]])

    [tex]e^{x} = ln(e^{e^5}}) [/tex]
    [tex]e^{x} = e^{e^5}}[/tex]
    [tex]e^{5} = (2.72)^{5}[/tex]
    [tex]e^{x} = e^{149}[/tex]
    [tex]x = 149 [/tex]

    ....I have no idea if I'm doing this right, but I'm not feeling like I am...Help?
     
  2. jcsd
  3. Jan 23, 2008 #2

    mjsd

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    are you trying to do two logs then two e's or one e then a log then two e's
    very confusing!
     
  4. Jan 23, 2008 #3
    You have to use one of the properties of logs. When the bases of the exponent and log are same, they cancel.
     
  5. Jan 23, 2008 #4

    Shooting Star

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    Hint: the answer is an integer.
     
  6. Feb 3, 2008 #5
    ln e ^x =x
    Can you take it from there?
     
  7. Feb 3, 2008 #6
    i think the answer is 5 since ln and e cancaled out!
     
  8. Feb 3, 2008 #7
    hey thanks everyone! i was able to figure it out from there you guys are always a big help :)
     
  9. Feb 9, 2008 #8
    This is true
     
  10. Feb 9, 2008 #9

    malawi_glenn

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    NEVER post the answer just like that!
     
  11. Feb 9, 2008 #10
    haha, I guess a perfect hint would be

    [tex] lne^x=x[/tex]
     
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