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Math Help - Log Help

  1. #1
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    Log Help

    Hi guys I have a few questions that I am totally lost on.

    1) Write as a single loagrithm: 4lnx - 2(lnx^3 + 4lnx)

    2) Solve: 2^x = 3^(x+3)

    3) Solve: log(x^2 -1) = 2 + log (x+1)

    Thanks in advance!!
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  2. #2
    Moo
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    Hello,

    Remember that \ln(a^b)=b \ln(a)

    So how would you transform \ln(x^3) in the first one ?

    For the second one, this is the same : compose each side with logarithm.

    For the third one, note that x-1=(x-1)(x+1)

    Then remember that \ln(ab)=\ln(a)+\ln(b)
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  3. #3
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    so for the first one would it go to:

    4lnx - 2(3lnx + 4lnx)

    and if so would it then get distributed to 4lnx - 6lnx + 8lnx?

    Sorry I'm really bad at this!!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Please do not double post antz
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by antz215 View Post
    so for the first one would it go to:

    4lnx - 2(3lnx + 4lnx)

    and if so would it then get distributed to 4lnx - 6lnx + 8lnx?

    Sorry I'm really bad at this!!
    no, it is 4ln(x) - 6ln(x) - 8ln(x)
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  6. #6
    Moo
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    Quote Originally Posted by antz215 View Post
    so for the first one would it go to:

    4lnx - 2(3lnx + 4lnx)

    and if so would it then get distributed to 4lnx - 6lnx + 8lnx?

    Sorry I'm really bad at this!!
    This is nearly it
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    Please do not double post antz
    Sorry! I wont do that again. Thank you both for your help!
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