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Logs homework help

  1. Sep 11, 2004 #1
    How do I solve for x in the interval [0,2pi]:
    2cos(x/3)=LnX


    Here's what I got:

    2(.5)=LnX
    1=LnX
    e^(1)=x
    x=2.718
    is this correct?


    If x=e^(Ln3)+Lne^2-5Ln1
    what is x?
    x=3+2-0=5
    is this correct?
     
  2. jcsd
  3. Sep 11, 2004 #2

    Tide

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    Was that supposed to be [itex]\cos \frac {\pi}{3}[/itex]?
     
  4. Sep 11, 2004 #3
    yes sorry, that is 2cos(pi/3)=LnX
     
    Last edited: Sep 11, 2004
  5. Sep 11, 2004 #4

    Tide

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    In that case, good job!
     
  6. Sep 11, 2004 #5
    how do I find the relative maximum value of the function y=(LnX)/X and the domain of the function Ln(X^2-1)?
     
  7. Sep 12, 2004 #6

    Tide

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    Just find the derivative of the function and set it to zero.
     
  8. Sep 12, 2004 #7
    for the domain or for the max value?
     
  9. Sep 12, 2004 #8

    Tide

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    That will give you the location of the relative maximum (you can verify that it is a maximum as opposed to a minimum). Then use that value of x that you find to determine the value of the function at that point.
     
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