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Math Help - Logarithim Question

  1. #1
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    Logarithim Question

    This is actually pretty urgent but for some reason I cant get into urgent Hw section. Been working on this for hours. And its for tomorrow.

    Ok question goes...

    Solve:

    Log (base3) (2-3x) = log (base9) (6x -19x +2)

    the answer is apparently -2 and -1/3. I can get -1/3 from solving a quadratic equation at the, but annoyinglyI get +2 as the other answer.

    So far I have :

    (lg (2-3x))/lg3 = (lg(6x -19x +2))/lg 9
    (lg 9) (lg [2-3x]) = (lg 3) (log [6x -19x +2])

    Im guessing this is where i go wrong because this is where the actual answers stop making sense

    9 (lg [2-3x]) = 3(lg [6x -19x +2])
    9(2-3x) = 3 (6x -19x +2)
    18-27x = 18x - 57x +6
    0 = 18x - 30x -12
    0 = (18x +6)(x-2)

    x=2, -1/3

    2 cannot be the answer beccause that would mean a negative log.

    Pleaseeee Help, it may easily be something silly because im very tired.

    Thanks,
    Owen
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bladeowen View Post
    This is actually pretty urgent but for some reason I cant get into urgent Hw section. Been working on this for hours. And its for tomorrow.

    Ok question goes...

    Solve:

    Log (base3) (2-3x) = log (base9) (6x -19x +2)

    the answer is apparently -2 and -1/3. I can get -1/3 from solving a quadratic equation at the, but annoyinglyI get +2 as the other answer.

    So far I have :

    (lg (2-3x))/lg3 = (lg(6x -19x +2))/lg 9
    (lg 9) (lg [2-3x]) = (lg 3) (log [6x -19x +2])

    Im guessing this is where i go wrong because this is where the actual answers stop making sense

    9 (lg [2-3x]) = 3(lg [6x -19x +2])
    9(2-3x) = 3 (6x -19x +2)
    18-27x = 18x - 57x +6
    0 = 18x - 30x -12
    0 = (18x +6)(x-2)

    x=2, -1/3

    2 cannot be the answer beccause that would mean a negative log.

    Pleaseeee Help, it may easily be something silly because im very tired.

    Thanks,
    Owen
    using the change of base formula was a good idea. here's a better one:

    note that if \log_a b = c, then \log_{\sqrt{a}} \sqrt{b} = c. that is, \log_a b = \log_{\sqrt{a}} \sqrt{b}.

    (try to prove that!)

    with that in mind, note that your problem simplifies to:

    \log_3 (2 - 3x) = \log_3 \sqrt{6x^2 - 19x + 2}

    \Rightarrow 2 - 3x = \sqrt{6x^2 - 19x + 2}

    \Rightarrow (2 - 3x)^2 = 6x^2 - 19x + 2

    i suppose you can take it from here

    (and, of course, be sure to test your answers in the original problem)
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  3. #3
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    Oct 2008
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    That is Brilliant

    Thank you so much.
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