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Properties of Ln and e^x)

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm just not sure what the answer to this is. I think it's an identity for e^x and ln, but I've never had a course that dealt with e^x or logs. So I don't know.

    What is the answer to e^14ln(x)? It's part of a larger problem, but I can't get the rest of it done until I know that.


    2. Relevant equations

    None.

    3. The attempt at a solution

    I think the answer is x^14. But I'm not sure.
     
  2. jcsd
  3. Sep 29, 2011 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, what was your reasoning?



    Aside: when you're writing linearly, be careful about parentheses! The expression
    e^14ln(x)​
    means
    [tex]e^{14} \ln(x)[/tex]
    whereas
    e^(14ln(x))​
    means
    [tex]e^{14 \ln(x)}[/tex]
     
  4. Sep 29, 2011 #3
    Yes, sorry about that. I did mean to write:

    e^(14*ln(x))

    I was thinking, that if I equated some random variable(say y) to e^(14ln(x)), then I could just solve that equation.

    y= e^(14*ln(x))
    lny= 14lnx

    lny = 14lnx

    lny = ln(x^14)

    e^ of both sides

    y = x^14


    The part of that I am unsure about is:

    Does 14lnx = ln(x^14) ?
     
  5. Sep 29, 2011 #4
    Yes, that works, so you can see that by definition, elnx = x. Also, a*lnx = ln(ax)
     
  6. Sep 29, 2011 #5

    Mark44

    Staff: Mentor

    That's not the best approach. The intended goal of this problem is for you to simplify the given expression. Setting your expression equal to, say, y, doesn't help much to move things toward your goal of simplification.

    Use the properties of logs and exponents to rewrite your expression in a different (and simpler) form.
    Yes.
     
  7. Sep 29, 2011 #6
    Thanks for the help!
     
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