1. Limited time only! Sign up for a free 30min personal 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!

Logarithm overkill!

  1. Jan 23, 2008 #1
    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:


    Here's what I've tried so far:


    [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
    use the property that Ln(a^b) = b Ln(a), Ln(e) = 1, Ln(e^a) = a
  4. Jan 23, 2008 #3


    User Avatar
    Homework Helper

    z=ln(lne^(e^5)). Use the definition of a log now.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Logarithm overkill!
  1. Logarithm's and Such (Replies: 4)

  2. Logarithm's and Such (Replies: 12)

  3. Finding the logarithm (Replies: 3)

  4. Proof on logarithms (Replies: 2)