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Math Help - Solve for x

  1. #1
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    Solve for x

    Solve for x:

    [log_3_(x)]^2 - 4log_9_(x) - 3 = 0


    just for reference,
    log_3_ means log base 3...


    thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    Solve for x:

    [log_3_(x)]^2 - 4log_9_(x) - 3 = 0


    just for reference,
    log_3_ means log base 3...


    thanks
    Hey Mr Green. just wanted to make sure it was [log_3_(x)]^2 and not log_3_(x)^2. as in its the whole log squared and not just the x
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  3. #3
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    thanks for making sure, but the way i posted it is correct. the whole log is squared, not only the x. so (log base 3 of x)Squared

    thanks.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jhevon View Post
    Hey Mr Green. just wanted to make sure it was [log_3_(x)]^2 and not log_3_(x)^2. as in its the whole log squared and not just the x
    Note that for some variable a:

    log_9_a = [sqrt(log_3_sqrt(a))]^2 = log_3_sqrt(3)

    In other words, if we have the log to some base of a number, it is equivalent to the log to the sqrt of the base of the squareroot of the number. How does this work? Here goes.

    [log_3_(x)]^2 - 4log_9_(x) - 3 = 0
    => [log_3_(x)]^2 - log_9_(x)^4 - 3 = 0
    => [log_3_(x)]^2 - log_3_(x)^2 - 3 = 0
    => [log_3_(x)]^2 - 2log_3_(x) - 3 = 0

    Let y be log_3_x
    => y^2 - 2y - 3 = 0
    => (y - 3)(y + 1) = 0
    => y = 3 or y = -1

    so log_3_x = 3 or log_3_x = -1
    => x = 27 or x = 1/3
    Last edited by Jhevon; March 4th 2007 at 11:00 AM.
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