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Math Help - Domain Logarithmic Function (Unsolved)

  1. #1
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    Thumbs up Domain Logarithmic Function (Solved)

    f(x)=-2ln(x+1)

    Find the domain of f.

    So, far this is I have:
    f(x)=-2ln(x+1)
    f(x)=-2e^(x+1)

    Can anyone go over this problem step by step with me?
    Last edited by wickwiki; April 6th 2008 at 08:17 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok to find the domain

    for f(x)=-2\ln(x+1)...the only problem is that x+1\geq{0} due to \ln(x)s properties...so x+1\geq{0},x\geq{-1}
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by wickwiki View Post
    f(x)=-2ln(x+1)

    Find the domain of f.

    So, far this is I have:
    f(x)=-2ln(x+1)
    f(x)=-2e^(x+1)

    Can anyone go over this problem step by step with me?
    The domain of

    g(x)=\ln(x)

    is the set  (0,\infty)

    so your function is related to the unshifted natrual log by

    f(x)=-2g(x+1)=-2\ln(x+1) This shifts the graph one unit left in the x direction so the domain is

    (-1,\infty)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    for f(x)=-2\ln(x+1)...the only problem is that x+1\geq{0} due to \ln(x)s properties...so x+1\geq{0},x\geq{-1}
    it must be strictly greater than zero. log(0) is undefined as well
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    O man

    I had it that way...but I was thinking of \sqrt{x} and I went back and changed it...on these forums I try to answer as fast as I can so I get messed up sometimes...thanks!...I knew that by the way =D
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ...I knew that by the way =D
    that's ok

    yes, i know you knew
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Haha

    ::blushes:: haha thanks
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  8. #8
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    Question

    OK, solution solved thanks for the help.

    I would like to know a bit more about how you solved it.

    f(x)=-2ln(x+1)
    So, what happens with the negative two exactly?

    Because the one of my similar problems I solved ln(x+4) and I got
    Domain= (-4,\infty).

    So, how is the -2 effect the equation?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    It doesnt

    The only thing that matters in a \ln(u(x)) is that the domain is restricted so that \forall{x}\in{Domain},u(x)>0...make sense?
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